12.1 The Arithmetic of Equations 12t1lara.weebly.com/uploads/1/6/3/2/1632178/ch12bookpdf.pdf ·...

Stoichiometry 353

12.1

FOCUSObjectives12.1.1 Explain how balanced equa-

tions apply to both chemistry and everyday situations.

12.1.2 Interpret balanced chemical equations in terms of moles, representative particles, mass, and gas volume at STP.

12.1.3 Identify the quantities that are always conserved in chemical reactions.

Guide for Reading

Build VocabularyParaphrase Introduce the term sto-ichiometry in your own words. Stress that stoichiometry allows students to calculate the amounts of chemical sub-stances involved in chemical reactions using information obtained from bal-anced chemical equations.

Reading StrategyRelate Text and Visuals Have stu-dents construct tables similar to Figure 12.3 to interpret balanced chemical equations for stoichiometric prob-lems. Students need to use only those levels of interpretation—particles, molecules, moles, mass, or gas vol-umes—appropriate to the problem.

INSTRUCT

Have students study the photograph and read the text that opens the sec-tion. Ask, How many cocoons would be required to produce enough silk for two Japanese kimonos? (twice as many, or 6000 cocoons) How did you calculate the number of cocoons? (by multiplying the number needed for one kimono by two)

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Answers to...Figure 12.1 Acceptable answers will list ingredients as reactants and cookies as the product.

Section ResourcesPrint• Guided Reading and Study Workbook,

Section 12.1• Core Teaching Resources, Section 12.1

Review• Transparencies, T122–T125• Laboratory Manual, Lab 19

Technology• Interactive Textbook with ChemASAP,

Problem Solving 12.1, 12.4, Assessment 12.1

• Go Online, Section 12.1

Connecting to Your World

Section 12.1 The Arithmetic of Equations 353

12.1 The Arithmetic of Equations

Silk is one of the most beautiful and luxurious of all fabrics. It is spun from the cocoons of silkworms. Silk

manufacturers know from experience that to pro-duce enough silk to make just one elegant Japa-

nese kimono they will need over 3000 cocoons. In a similar fashion, chemists need to know how much reactant is needed to make a certain amount of product. The

answer lies in chemical equations. From a balanced chemical equation, you can deter-

mine the quantities of reactants and products in a reaction.

Guide for Reading

Key Concepts• How is a balanced equation like

a recipe?• How do chemists use balanced

chemical equations?• In terms of what quantities can

you interpret a balanced chemical equation?

• What quantities are conserved in every chemical reaction?

Vocabularystoichiometry

Reading StrategyUsing Prior Knowledge Before you read, jot down three things you know about balanced chemi-cal equations. When you have read the section, explain how what you already knew helped you learn something new.

Using Everyday EquationsWhen you bake cookies, you probably use a recipe. A cookie recipe tells youthe precise amounts of ingredients to mix to make a certain number ofcookies, as shown in Figure 12.1. If you need a larger number of cookiesthan the recipe provides, you can double or triple the amounts of ingredi-ents. A balanced chemical equation provides the same kind of quantita-tive information that a recipe does. In a cookie recipe, you can think of theingredients as the reactants, and the cookies as the products.

Here is another example. Imagine you are in charge of manufacturing forthe Tiny Tyke Tricycle Company. The business plan for Tiny Tyke requires theproduction of 640 custom-made tricycles each week. One of your responsibi-lities is to be sure that there are enough parts available at the start of eachworkweek to make these tricycles. How can you determine the number ofparts you need per week?

Figure 12.1 A cookie recipe tells you the number of cookies that you can expect to make from the listed amounts of ingredients. Using ModelsHow can you express a cookie recipe as a balanced equation?

354 Chapter 12

Section 12.1 (continued)

Using Everyday EquationsDiscussWrite this statement on the board: “A frame, a seat, wheels, a handlebar, and pedals are needed to assemble a com-plete tricycle.” Have students look at the visual representation of how a tri-cycle is assembled and ask, Does this statement correctly describe the process? (no) What is the correct statement? (A frame, a seat, three wheels, a handlebar, and two pedals are needed to assemble a complete tricycle.)

Using Balanced Chemical EquationsDiscussRemind students that during a chemi-cal reaction, atoms are rearranged into new combinations and groupings. It is somewhat similar to changing silk cocoons into a kimono. Just as a chemical reaction has reactants and products, the cocoon is the reactant and the kimono is the product. Much like having the proper supply of silk cocoons, chemists must have an adequate supply of reactants for a chemical reaction.

Word OriginsThe term spectrometry means “the study of light or spectrum.”

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354 Chapter 12

Word Origins

To simplify this discussion, assume that the major components of thetricycle are the frame (F), the seat (S), the wheels (W), the handlebars (H),and the pedals (P), in other words, your reactants. The figure below illus-trates how an equation can represent the manufacturing of a single tricycle.

The finished tricycle, your product, has a “formula” of FSW3HP2. Thebalanced equation for making a single tricycle is

F � S � 3W � H � 2P ¡ FSW3HP2

This balanced equation is a “recipe” to make a single tricycle: Making a tri-cycle requires assembling one frame, one seat, three wheels, one handle-bar, and two pedals. Now look at Sample Problem 12.1. It shows you how touse the balanced equation to calculate the number of parts needed to man-ufacture a given number of tricycles.

Using Balanced Chemical EquationsNearly everything you use is manufactured from chemicals—soaps, sham-poos and conditioners, CDs, cosmetics, medicines, and clothes. In manu-facturing such items, the cost of making them cannot be greater than theprice they are sold at. Otherwise, the manufacturer will not make a profit.Therefore, the chemical processes used in manufacturing must be carriedout economically. This is where balanced equations help.

A balanced chemical equation tells you what amounts of reactants tomix and what amounts of product to expect. Chemists use balancedchemical equations as a basis to calculate how much reactant is needed orproduct is formed in a reaction. When you know the quantity of one sub-stance in a reaction, you can calculate the quantity of any other substanceconsumed or created in the reaction. Quantity usually means the amountof a substance expressed in grams or moles. However, quantity could justas well be in liters, tons, or molecules.

The calculation of quantities in chemical reactions is a subject ofchemistry called stoichiometry. Calculations using balanced equations arecalled stoichiometric calculations. For chemists, stoichiometry is a form ofbookkeeping. For example, accountants can track income, expenditures,and profits for a small business by tallying each in dollars and cents. Chem-ists can track reactants and products in a reaction by stoichiometry. Itallows chemists to tally the amounts of reactants and products using ratiosof moles or representative particles.

Checkpoint How is stoichiometry similar to bookkeeping?

Stoichiometry comes from the combination of the Greek words stoikheioin,meaning “element,” and metron, meaning “to mea-sure.” Stoichiometry is the calculation of amounts of substances involved in chemical reactions. What do think the term spec-trometry means?

F � S � 3W � �H 2P FSW3HP2

Stoichiometry 355

DiscussWrite the equation that represents the production of a tricycle.

F + S + 3W + H + 2P → FSW3HP2

Point out that the balanced equation contains information that not only relates reactant to products but relates one reactants to another.

Ask, How many handlebars are needed to complete tricycles if 24 pedals are available? (12) Show stu-dents how to set up the proportions to solve this problem.

Ask, How many handlebars are needed to complete tricycles if 24 seats are available? (24) 24 wheels? (8)

Sample Problem 12.1

Answers1. 288 seats, 864 wheels, 576 pedals2. Answers will vary but should

include the correct number of “parts” to make the product.

Practice Problems PlusChapter 21 Assessment problem 57 is related to Sample Problem 12.1.

Math HandbookFor a math refresher and preview, direct students to dimensional analysis, page R66.

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? H = 24 P � (1 H/2P) = 12 H

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Answers to...

Checkpoint

In stoichiome-try, the amount of product formed by a chemical reaction can be calcu-lated based on the amounts of the reactants. In bookkeeping, the prof-its of a business can be calculated based on income and expenditures.


Practice Problems

Practice Problems

SAMPLE PROBLEM 12.1

Using a Balanced Equation as a RecipeIn a five-day workweek, Tiny Tyke is scheduled to make 640 tricycles.How many wheels should be in the plant on Monday morning to makethese tricycles?

Analyze List the knowns and the unknown.

Knowns• number of tricycles � 640 tricycles � 640 FSW3HP2

• F � S � 3W � H � 2P ¡ FSW3HP2

Unknown• number of wheels � ? wheels

The desired conversion is tricycles (FSW3HP2) ¡ wheels (W). Thebalanced equation tells you that each tricycle has three wheels, or1 FSW3HP2 � 3 W. The problem can be solved by using the properconversion factor derived from this expression.

Calculate Solve for the unknown.

You can write two conversion factors relating wheels to tricycles.

The desired unit is W, so use the conversion factor on the left—the onethat has W in the numerator. Multiply the number of tricycles by theconversion factor.

Evaluate Does the result make sense?

If three wheels are required for each tricycle, and a total of more than600 tricycles are being made, then a number of wheels in excess of1800 is a logical answer. The unit of the known cancels with the unit inthe denominator of the conversion factor, and the answer is in the unitof the unknown.

3 W1 FSW3HP2

and 1 FSW3HP2

3 W

640 FSW3HP2 � 3W1 FSW3HP2

� 1920 W

1. Tiny Tike has decided to make 288 tricycles each day. How many tricycle seats, wheels, and pedals are needed?

2. Write an equation that gives your own “recipe” for making a skateboard.

Math Handbook

For help with dimensional analysis, go to page R66.

withChemASAP

Problem Solving 12.1 Solve Problem 1 with the help of an interactive guided tutorial.

356 Chapter 12


Interpreting Chemical EquationsDiscussReview balancing chemical reactions by writing several unbalanced equa-tions on the board. For example:

CuO(s) +NH3(aq) → Cu(s) + H2O(l) + N2(g)

NH3(g) + O2(g) → NO(g) + H2O(g)

KClO3(s) → KCl(s) + O2(g).

Have students balance the equations as shown.

3CuO(s) + 2NH3(aq) → 3Cu(s) + 3H2O(l) + N2(g)

4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g)

2KClO3(s) → 2KCl(s) + 3O2(g)

Ask, Why is it not correct to balance an equation by changing the sub-scripts in one or more of the formu-las? (Changing the subscripts in a formula changes the chemical identity of the substance.)

Mass Conservation in Chemical ReactionsUse VisualsFigure 12.3 Remind students that the term STP represents “standard temper-ature and pressure.” Ask, What are the values of STP? (0°C and 101.3 kPa) Why is the volume of a gas usually measured at STP? (The volume of a gas is usually measured at STP because its volume varies with temperature and pressure.) What is the molar volume of any gas at STP? (22.4 L/mol) How many particles does it contain? (22.4 L of any ideal gas at STP contains 6.02 × 1023 particles of that gas.)

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Ammonia in the Nitrogen CycleEarth’s atmosphere contains 0.01 parts per million of ammonia, and small amounts of ammonia occur in volcanic gases. Most ammonia cycles through the living world without returning to the atmosphere. Ammonia plays a role in several stages of the nitrogen cycle. Nitrogen-fixing bacteria form nodules, or swellings, on the roots of plants in the legume family, such as beans and

clover plants. These bacteria change atmo-spheric nitrogen into ammonia molecules or ammonium ions. Other bacteria break down the nitrogenous material in dead plants and animals into ammonia molecules. Certain soil bacteria oxidize these molecules into nitrate ions, the form readily absorbed by plant roots. When a plant dies, this cycle begins again.

Facts and Figures

356 Chapter 12

Figure 12.2 Gardeners use ammonium salts as fertilizer. The nitrogen in these salts is essential to plant growth.

Interpreting Chemical EquationsIn gardens such as the one shown in Figure 12.2, fertilizers are often used toimprove the growth of flowers. As you may recall from Chapter 10, ammo-nia is widely used as a fertilizer. Ammonia is produced industrially by thereaction of nitrogen with hydrogen.

N2(g) � 3H2(g) ¡ 2 NH3(g)

The balanced chemical equation tells you the relative amounts of reac-tants and product in the reaction. However, your interpretation of theequation depends on how you quantify the reactants and products. Abalanced chemical equation can be interpreted in terms of different quanti-ties, including numbers of atoms, molecules, or moles; mass; and volume. Asyou study stoichiometry, you will learn how to read a chemical equation interms of any of these quantities.

Number of Atoms At the atomic level, a balanced equation indicatesthat the number and type of each atom that makes up each reactant alsomakes up each product. Thus, both the number and types of atoms are notchanged in a reaction. In the synthesis of ammonia, the reactants are com-posed of two atoms of nitrogen and six atoms of hydrogen. These eightatoms are recombined in the product.

Number of Molecules The balanced equation indicates that one mol-ecule of nitrogen reacts with three molecules of hydrogen. Nitrogen andhydrogen will always react to form ammonia in a 1:3:2 ratio of molecules. Ifyou could make 10 molecules of nitrogen react with 30 molecules of hydro-gen, you would expect to get 20 molecules of ammonia. Of course, it is notpractical to count such small numbers of molecules and allow them toreact. You could, however, take Avogadro’s number of nitrogen moleculesand make them react with three times Avogadro’s number of hydrogen mol-ecules. This would be the same 1:3 ratio of molecules of reactants. The reac-tion would form two times Avogadro’s number of ammonia molecules.

Moles A balanced chemical equation also tells you the number of moles ofreactants and products. The coefficients of a balanced chemical equationindicate the relative numbers of moles of reactants and products in a chemi-cal reaction. This is the most important information that a balanced chemi-cal equation provides. Using this information, you can calculate the amountsof reactants and products. In the synthesis of ammonia, one mole of nitrogenmolecules reacts with three moles of hydrogen molecules to form two molesof ammonia molecules. As you can see from this reaction, the total numberof moles of reactants does not equal the total number of moles of product.

Checkpoint What do the coefficients of a balanced chemical equation indicate?

Stoichiometry 357

Download a worksheet on Conservation of Mass for students to complete and find additional support for NSTA SciLinks.

TEACHER DemoTEACHER Demo

Interpreting a Chemical EquationPurpose Students interpret a bal-anced equation of the reaction of mag-nesium and hydrochloric acid.

Materials 2.5–3.5-cm strip of magne-sium, 50 mL 1M HCl(aq) in a 100-mL beaker, baking soda

Safety Wear safety glasses and an apron. Neutralize remaining HCl(aq) with baking soda before flushing down the drain.

Procedure Identify the two reactants as magnesium and hydrochloric acid. Have students observe the reaction as you carefully add the magnesium strip to the acid. Ask students to write a bal-anced chemical equation for the reac-tion of magnesium and hydrochloric acid. [Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)] Have students interpret the equation in terms of particles, moles, and molar masses.

Expected Outcome Students should express the balanced equation at the particle level as one atom of magne-sium reacts with two molecules of hydrogen chloride to produce one for-mula unit of magnesium chloride and one molecule of hydrogen gas. Simi-larly, one mole of magnesium reacts with two moles of hydrogen chloride to produce one mole of magnesium chloride and one mole of hydrogen gas. Finally, 24.31 g Mg + 72.92 g HCl produces 95.21 g MgCl2 + 2.02 g H2.

Answers to... Figure 12.3 10 molecules NH3

Checkpoint

the relative numbers of representative particles, the relative numbers of moles, and, for gases, the relative volumes

Less Proficient Readers Have students construct a table like the one shown in Figure 12.3 for the reaction of hydrogen gas with oxygen gas to form water. Students should begin by writing the bal-anced equation. Encourage students to draw pictures as shown in Figure 12.3 to represent reactants and products. Have students use

the completed table to answer questions such as, “How many moles of water are produced by reacting 4 moles of hydrogen gas with excess oxygen?” Other reactions for students to analyze:CO(g) + 2H2(g) → CH3OH(l) CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

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Differentiated Instruction


Mass A balanced chemical equation obeys the law of conservation ofmass. This law states that mass can be neither created nor destroyed in anordinary chemical or physical process. As you recall, the number and typeof atoms does not change in a chemical reaction. Therefore, the total massof the atoms in the reaction does not change. Using the mole relationship,you can relate mass to the number of atoms in the chemical equation. Themass of 1 mol of N2 (28.0 g) plus the mass of 3 mol of H2 (6.0 g) equals themass of 2 mol of NH3 (34 g). Although the number of moles of reactantsdoes not equal the number of moles of product, the total number of gramsof reactants does equal the total number of grams of product.

Volume If you assume standard temperature and pressure, the equationalso tells you about the volumes of gases. Recall that 1 mol of any gas at STPoccupies a volume of 22.4 L. The equation indicates that 22.4 L of N2 reactswith 67.2 L (3 � 22.4 L) of H2. This reaction forms 44.8 L (2 � 22.4 L) of NH3.

Mass Conservation in Chemical ReactionsFigure 12.3 summarizes the information derived from the balanced chemi-cal equation for the formation of ammonia. As you can see, the mass of thereactants equals the mass of the products. In addition, the number ofatoms of each type in the reactants equals the number atoms of each typein the product. Mass and atoms are conserved in every chemicalreaction. However, molecules, formula units, moles, and volumes are notnecessarily conserved—although they may be. Consider, for example, theformation of hydrogen iodide,

H2(g) � I2(g) ¡ 2HI(g)

In this reaction, molecules, moles, and volume are all conserved. But in themajority of chemical reactions, they are not..

Figure 12.3 The balanced chemical equation for the formation of ammonia can be interpreted in several ways.Predicting How many mol-ecules of NH3 could be made from 5 molecules of N2 and 15 molecules of H2?1 molecule N2

10 molecules N2

1 mol N2

28 g N2

2 atoms N

N2(g)

1 �6.02 � 1023

molecules N2( )

22.4 L N2

AssumeSTP

2 molecules NH3

20 molecules NH3

2 mol NH3

2 � 17 g NH3

34 g products

2 atoms N and6 atoms H

2NH3(g)

2�6.02 � 1023

molecules NH3

44.8 L NH3

3 molecules H2

30 molecules H2

3 mol H2

3 � 2 g H2

34 g reactants

6 atoms H

3H2(g)

3 �6.02 � 1023

molecules H2( )

67.2 L H2

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( )1 molecule N2

10 molecules N2

1 mol N2

28 g N2

2 atoms N

N2(g)

1 �6.02 � 1023

molecules N2( )

22.4 L N2

AssumeSTP

2 molecules NH3

20 molecules NH3

2 mol NH3

2 � 17 g NH3

34 g products

2 atoms N and6 atoms H

2NH3(g)

2�6.02 � 1023

molecules NH3

44.8 L NH3

3 molecules H2

30 molecules H2

3 mol H2

3 � 2 g H2

34 g reactants

6 atoms H

3H2(g)

3 �6.02 � 1023

molecules H2( )

67.2 L H2

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For: Links on Conservation of Mass

Visit: www.SciLinks.orgWeb Code: cdn-1121

358 Chapter 12


CONCEPTUAL PROBLEM 12.1

Answers3. 2 molecules H2 + 1 molecule O2→

2 molecules H2O2 mol H2 + 1 mol O2 → 2 mol H2O 44.8 L H2 + 22.4 L O2 → 44.8 L H2O

4. 2 mol C2H2 + 5 mol O2 → 4 mol CO2 + 2 mol H2O

44.8 L C2H2 + 112 L O2 → 89.6 L CO2 + 44.8 L H2O

212 g reactants → 212 g products

Practice Problems PlusBalance the following equation:

C5H12(g) + O2(g) → CO2(g) + H2O(g)

Interpret the balanced equation in terms of relative number of moles, volumes of gas at STP, and masses of reactants and products. [1 mol C5H12(g) + 8 mol O2(g) → 5 mol CO2(g) + 6 mol H2O(g); 22.4 L C5H12(g) + 179 L O2(g) → 112 L CO2(g) + 134 L H2O(g); 328 g reactants → 328 g products]

ASSESSEvaluate UnderstandingHave pairs of students write a balanced chemical equation for a simple reaction. Have pairs exchange equations and write quantitative relationships between reactants and products in terms of mass, moles, particles, and, where appropriate, volumes.

ReteachRemind students that the coefficients in a balanced chemical equation state the relationships among substances involved in the reaction.

Moles of reactants and products are conserved for some reactions, but this is generally not the case.

with ChemASAP

If your class subscribes to the Interactive Textbook, use it to review key concepts in Section 12.1.

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Section 12.1 Assessment5. Both a balanced equation and a recipe

give quantitative information about the starting and end materials.

6. as a basis to calculate how much reactant is needed or product is formed in a reac-tion

7. numbers of atoms, molecules, or moles; mass; and volumes

8. mass and atoms

9. 2 atoms K + 2 molecules H2O → 2 formula units KOH + 1 molecule H2

2 mol K + 2 mol H2O → 2 mol KOH + 1 mol H2

78.2 g K + 36.0 g H2O → 112.2 g KOH + 2.0 g H2

10. C2H5OH + 3O2 → 2CO2 + 3H2O46.0 g C2H5OH + 96.0 g O2 →

88.0 g CO2 + 54.0 g H2O142.0 g reactants → 142.0 g products

358 Chapter 12

Practice Problems

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withChemASAP

3. Interpret the equation for the formation of water from its elements in terms of numbers of mole-cules and moles, and volumes of gases at STP.

2H2(g) � O2(g) ¡ 2H2O(g)

4. Balance the following equation.

C2H2(g) � O2(g) ¡ CO2(g) � H2O(g)

Interpret the balanced equation in terms of

relative numbers of moles, volumes of gas at STP, and masses of reactants and products.

Analyze Identify the relevant concepts.

a. The coefficients in the balanced equation give the relative number of molecules or moles of reactants and products.

b. A balanced chemical equation obeys the law of conservation of mass.

Solve Apply concepts to this situation.

a. 2 molecules H2S � 3 molecules O2 ¡2 molecules SO2 � 2 molecules H2O

2 mol H2S � 3 mol O2 ¡2 mol SO2 � 2 mol H2O

b. Multiply the number of moles of each reactant and product by its molar mass:

2 mol H2S � 3 mol O2 ¡ 2 mol SO2 � 2 mol H2O.

68.2 g H2S � 96.0 g O2 ¡ 128.2 g SO2 � 36.0 g H2O

164.2 g � 164.2 g

a2 mol � 34.1g

molb + a3 mol � 32.0g

molb ¡

a2 mol � 64.1g

molb + a2 mol � 18.0g

molb

withChemASAP

Problem-Solving 12.4 Solve Problem 4 with the help of an interactive guided tutorial.

12.1 Section Assessment

5. Key Concept How is a balanced equation simi-lar to a recipe?

6. Key Concept How do chemists use balanced equations?

7. Key Concept Chemical reactions can be described in terms of what quantities?

8. Key Concept What quantities are always con-served in chemical reactions?

9. Interpret the given equation in terms of relative numbers of representative particles, numbers of moles, and masses of reactants and products.

2K(s) � 2H2O(l) ¡ 2KOH(aq) � H2(g)

10. Balance this equation: C2H5OH(l) � O2(g) ¡CO2(g) � H2O(g). Show that the balanced equation obeys the law of conservation of mass.

Explanatory Paragraph Explain this statement: “Mass and atoms are conserved in every chemical reaction, but moles are not necessarily conserved.”

Assessment 12.1 Test yourself on the concepts in Section 12.1.

CONCEPTUAL PROBLEM 12.1

Interpreting a Balanced Chemical EquationHydrogen sulfide, which smells like rotten eggs, is found in volcanicgases. The balanced equation for the burning of hydrogen sulfide is:

2H2S(g) � 3O2(g) ¡ 2SO2(g) � 2H2O(g)

Interpret this equation in terms ofa. numbers of representative particles and moles.b. masses of reactants and products.

Practice Problems

Stoichiometry 359

Print• Guided Reading and Study Workbook,

Section 12.2• Core Teaching Resources,

Section 12.2 Review• Transparencies, T126–T132• Laboratory Manual, Lab 19

• Small-Scale Chemistry Laboratory Manual, Labs 18, 19

• Probeware Laboratory Manual, Section 12.2


Simulation 13, Problem-Solving 12.12, 12.13, 12.15, 12.19, Assessment 12.2

12.2

FOCUSObjectives12.2.1 Construct mole ratios from bal-

anced chemical equations and apply these ratios in stoichio-metric calculations.

12.2.2 Calculate stoichiometric quan-tities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP.

Guide for Reading

Build VocabularyParaphrase Have students work with a partner to define the term mole ratio in their own words. They may do so by reading this section and by using what they have already learned about bal-anced chemical equations. Have stu-dent pairs read their definitions to the class.

Reading StrategyIdentify Main Ideas/Details As you read the material under the heading Mass-Mass Calculations, identify and list the main ideas presented by the text.

INSTRUCT

Have students study the photograph and read the text that opens the sec-tion. Write the equation for the decom-position of sodium azide with heat as one of the products 2NaN3(s) → 2Na(s) + 3N2(g) + heat. Ask, How can stoichi-ometry be used to calculate the vol-ume of a gas produced in this reaction? (The number of moles (and volume) of nitrogen gas formed by this reaction depends on the number of moles of sodium azide that decompose.)

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Section Resources

Connecting to Your World

Section 12.2 Chemical Calculations 359

Writing and Using Mole RatiosAs you just learned, a balanced chemical equation provides a great deal ofquantitative information. It relates particles (atoms, molecules, formulaunits), moles of substances, and masses. A balanced chemical equationalso is essential for all calculations involving amounts of reactants andproducts. For example, suppose you know the number of moles of one sub-stance. The balanced chemical equation allows you to determine the num-ber of moles of all other substances in the reaction.

Look again at the balanced equation for the production of ammoniafrom nitrogen and hydrogen:

N2(g) � 3H2(g)¡ 2NH3(g)

The most important interpretation of this equation is that 1 mol of nitrogenreacts with 3 mol of hydrogen to form 2 mol of ammonia. Based on thisinterpretation, you can write ratios that relate moles of reactants to molesof product. A mole ratio is a conversion factor derived from the coefficientsof a balanced chemical equation interpreted in terms of moles. Inchemical calculations, mole ratios are used to convert between moles of reac-tant and moles of product, between moles of reactants, or between moles ofproducts. Three mole ratios derived from the balanced equation above are:

Mole-Mole Calculations In the mole ratio below, W is the unknownquantity. The values of a and b are the coefficients from the balanced equa-tion. Thus a general solution for a mole-mole problem, such as SampleProblem 12.2, is given by

1 mol N2

3 mol H2

2 mol NH3

1 mol N2

3 mol H2

2 mol NH3

x mol G � b mol Wa mol G � xb

a mol W

Given Mole ratio Calculated

12.2 Chemical Calculations

Guide for Reading

Key Concepts• How are mole ratios used in

chemical calculations?• What is the general procedure

for solving a stoichiometric problem?

Vocabularymole ratio

Reading StrategyRelating Text and Visuals As you read, look closely at Figure 12.9. Explain how this illustration helps you understand the rela-tionship between known and unknown quantities in a stoichio-metric problem.

Figure 12.4 Manufacturingplants produce ammonia by combining nitrogen with hydrogen. Ammonia is used in cleaning products, fertilizers, and in the manufacture of other chemicals.

Air bags inflate almost instanta-neously upon a car’s impact. The effectiveness of air bags is based on the rapid conversion of a small mass of sodium azide into a large volume of

gas. The gas fills an air bag, preventing the driver from hitting the steering wheel

or dashboard. The entire reaction occurs in less than a second. In this section you will learn how to use a balanced chemical equation to

calculate the amount of product formed in a chemical reaction.

360 Chapter 12


Writing and Using Mole RatiosUsing Visuals Figure 12.4 Have students consider the sizes of the containers shown rela-tive to the sizes of containers used in a classroom laboratory. Have them imag-ine they are managing the manufac-turing facility pictured. Ask, What factors would they need to consider to meet demands for ammonia? (Acceptable answers include the number of customers, the number of cylinders per customer, the amount of ammonia per cylinder, and the amount of H2 and N2 needed to produce that quantity of NH3.)

Sample Problem 12.2

Answers11. a.

b. 7.4 mol12. a. 11.1 mol

b. 0.52 mol


Math HandbookFor a math refresher and practice, direct students to dimensional anal-ysis, page R66.

DiscussPoint out that heat is produced in the decomposition of sodium azide, used in air safety bags. Explain the heat pro-duced by a reaction can also be mea-sured and related to the amount of reactant(s) consumed. In this case, the amount of heat produced depends on the mass of sodium azide (the reac-tant) that decomposes. It is possible to relate grams of reactant to moles of reactant to heat produced.

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4 mol Al

3 mol O2

4 mol AI

2 mol AI2O3

3 mol O2

4 mol AI

2 mol AI2O3

3 mol O2

3 mol O2

2 mol AI2O3

2 mol AI2O3

4 mol AI

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Less Proficient Readers Encourage students to find a method of problem solving that capitalizes on their strengths. For example, a visual learner might draw pictures of the reactants and products, instead of just writing the sym-bols. A kinesthetic learner may prefer to manipulate molecular models of the reac-tants and products.

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360 Chapter 12

Practice Problems

Practice Problems

SAMPLE PROBLEM 12.2

Calculating Moles of a ProductHow many moles of ammonia are produced when 0.60 mol of nitrogenreacts with hydrogen?

Analyze List the known and the unknown.

Known• moles of nitrogen � 0.60 mol N2

Unknown• moles of ammonia � ? mol NH3

The conversion is mol N2 ¡ mol NH3. According to the balancedequation, 1 mol N2 combines with 3 mol H2 to produce 2 mol NH3. Todetermine the number of moles of NH3, the given quantity of N2 ismultiplied by the form of the mole ratio from the balanced equation thatallows the given unit to cancel. This mole ratio is 2 mol NH3/1 mol N2.



The ratio of 1.2 mol NH3 to 0.60 mol N2 is 2:1, as predicted by thebalanced equation.

0.60 mol N2 �2 mol NH3

1 mol N2� 1.2 mol NH3

11. This equation shows the forma-tion of aluminum oxide, which is found on the surface of alumi-num objects exposed to the air.

4Al(s) � 3O2(g)¡ 2Al2O3(s)

a. Write the six mole ratios that can be derived from this equation.

b. How many moles of alumi-num are needed to form 3.7 mol Al2O3?

12. According to the equation in Problem 11:a. How many moles of oxygen

are required to react com-pletely with 14.8 mol Al?

b. How many moles of Al2O3

are formed when 0.78 mol O2 reacts with aluminum?

Mass-Mass Calculations No laboratory balance can measure substa-nces directly in moles. Instead, the amount of a substance is usually deter-mined by measuring its mass in grams, as shown in Figure 12.5. From themass of a reactant or product, the mass of any other reactant or product in agiven chemical equation can be calculated. The mole interpretation of a bal-anced equation is the basis for this conversion. If the given sample is mea-sured in grams, the mass can be converted to moles by using the molarmass. Then the mole ratio from the balanced equation can be used to calcu-late the number of moles of the unknown. If it is the mass of the unknownthat needs to be determined, the number of moles of the unknown can bemultiplied by the molar mass. As in mole-mole calculations, the unknowncan be either a reactant or a product.

Figure 12.5 To determine the number of moles in a sample of a compound, first measure the mass of the sample. Then use the molar mass to calculate the number of moles in that mass.

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Stoichiometry 361

Sample Problem 12.3

Answers13. 2.03 g C2H214. 1.36 mol CaC2

Practice Problems PlusRust (Fe2O3) is produced when iron (Fe) reacts with oxygen (O2). 4Fe(s) + 3O2(g) → 2Fe2O3(s)How many grams of Fe2O3 are pro-duced when 12.0 g of iron rusts? (17.2 g)

Significant FiguresPoint out the concept of signifi-cant figures applies only to mea-sured quantities. Review the following condensed rules for determining which digits are significant.1. Nonzero digits are significant.2. a. A zero is significant only if it

as at the right end of a number and after a decimal point, orb. between digits that are signifi-cant according to rules 1 and 2a.

3. If a quantity is exact, it has an unlimited number of significant digits.

4. All digits of a quantity written in scientific notation are signifi-cant. Ask, How many signifi-cant figures are in each of the following measurements?

Math HandbookFor a math refresher and practice, direct students to significant fig-ures, page R59.

a. 30, 400 s (3)

b. 150.0 cm (4)

c. 2401 km (4)

d. 168.030 m (6)

e. 0.058 m (2)

f. 3.010 × 108 s (4)

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Practice Problems

Practice Problems

SAMPLE PROBLEM 12.3

Calculating the Mass of a ProductCalculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. The balanced equation is

N2(g) � 3H2(g)¡ 2NH3(g)


Knowns• mass of hydrogen � 5.40 g H2

• 3 mol H2 � 2 mol NH3 (from balanced equation)• 1 mol H2 � 2.0 g H2 (molar mass)• 1 mol NH3 � 17.0 g NH3 (molar mass)Unknown• mass of ammonia � ? g NH3

The mass in grams of hydrogen will be used to find the mass in gramsof ammonia:

g H2¡ g NH3

The following steps are necessary to determine the mass of ammonia:

g H2¡mol H2¡mol NH3¡ g NH3

The coefficients of the balanced equation show that 3 mol H2 reactswith 1 mol N2 to produce 2 mol NH3. The mole ratio relating mol NH3

to mol H2 is 2 mol NH3/3 mol H2.


This following series of calculations can be combined:

g H2¡mol H2¡mol NH3¡ g NH3


Because there are three conversion factors involved in this solution, itis more difficult to estimate an answer. However, because the molarmass of NH3 is substantially greater than the molar mass of H2, theanswer should have a larger mass than the given mass. The answershould have two significant figures.

Given Change given Mole ratio Change molesquantity unit to moles to grams

5.40 g H2 �1 mol H2

2.0 g H2�

2 mol NH3

3 mol H2�

17.0 g NH3

1 mol NH3� 31 g NH3

Significant FiguresThe significant figures in a measurement are all the dig-its known with certainty plus one estimated digit. The number of significant figures in the measurements used in a calculation determines how you round the answer.

When multiplying and dividing measurements, the rounded answer can have no more significant figures than the least number of signifi-cant figures in any measure-ment in the calculation.

The product of 3.6 m �2.48 m � 8.928 m2 is rounded to 8.9 m2 (2 signifi-cant figures).

When adding and sub-tracting measurements, the answer can have no more decimal places than the least number of decimal places in any measurement in the problem. The difference of 8.78 cm � 2.2 cm � 6.58 cm is rounded to 6.6 cm (one decimal place).

13. Acetylene gas (C2H2) is pro-duced by adding water to calcium carbide (CaC2).CaC2(s) � 2H2O(l)¡

C2H2(g) � Ca(OH)2(aq)

How many grams of acety-lene are produced by adding water to 5.00 g CaC2?

14. Using the same equation, deter-mine how many moles of CaC2

are needed to react completely with 49.0 g H2O.


Math Handbook

For help with significant figures, go to page R59.

362 Chapter 12



Interpreting a Chemical EquationPurpose Students interpret a balanced equation in terms of moles and mass.

Materials Prior to the demonstration prepare 0.1M solutions of potassium iodide and lead(II) nitrate. Measure 50.0 mL of Pb(NO3)2 and 150 mL of KI into separate 250-mL beakers.

Safety Wear safety glasses and apron.

Procedure Tell students that you are going to mix 0.005 moles of lead(II) nitrate with excess potassium iodide. Have student observe as you combine both solutions in the 250-mL beaker. Have students write a balanced chemi-cal equation for the observed reaction.

[2KI(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbI2(s)]

Have students predict the number of moles of product produced. (0.005 moles PbI2 assuming the reaction was complete) Note that, in an actual reac-tion, the amounts of reactants often are not present in the mole ratios predicted by the coefficients in a balanced equa-tion. Explain the importance of the mole ratios in an equation for calculat-ing relative quantities. Ask, What is the mass of lead(II) nitrate reacted and the mass of lead(II) iodide produced? (1.66 g Pb(NO3)2 and 2.30 g PbI2)

Expected Outcome A bright yellow precipitate will form.

DiscussStudents sometimes try to do mass-mass conversions by incorrectly using the mole ratio as a mass ratio. (That is, they use grams instead of moles as the units in the mole ratio, and then skip the mass-mole conversion step.) Stress that because the number of grams in one mole of a substance varies with its molar mass, a mass-mole conversion is a necessary intermediate step in mass-mass stoichiometric problems.

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Atmospheric AmmoniaAmmonia is found in trace amounts in the atmospheres of three Jovian planets—Jupiter, Saturn, and Uranus. In Jupiter’s atmosphere, the clouds of ammonia consist of frozen ammonia droplets changing to liquid

ammonia droplets nearer the planet’s surface. Because of colder temperatures, the ammonia clouds in the atmosphere of Saturn and Uranus consist of frozen ammonia droplets.

362 Chapter 12

(given quantity) (wanted quantity)

aG bW

Mass-moleconversion

Mole ratio frombalanced equation

Mole-massconversion

�mass

of G

1 mol Gmass G

mol G �b mol Wa mol G

mol W �mass W1 mol W

mass

of W

If the law of conservation of mass is true, how is it possible to make 31 gNH3 from only 5.40 g H2? Looking back at the equation for the reaction, youwill see that hydrogen is not the only reactant. Another reactant, nitrogen, isalso involved. If you were to calculate the number of grams of nitrogenneeded to produce 31 g NH3 and then compare the total masses of reactantsand products, you would have an answer to this question. Go ahead and try it!

Mass-mass problems are solved in basically the same way as mole-mole problems. Figure 12.7 reviews the steps for the mass-mass conversionof any given mass (G) and any wanted mass (W ).

Figure 12.6 In this Hubble Space Telescope image, clouds of condensed ammonia are visible covering the surface of Saturn.

Figure 12.7 This general solution diagram indicates the steps necessary to solve a mass-mass stoichiometry problem: convert mass to moles, use the mole ratio, and then convert moles to mass. Inferring Is the given always a reactant?

Steps in Solving a Mass-Mass Problem

1. Change the mass of G to moles of G (mass G ¡ mol G) byusing the molar mass of G.

2. Change the moles of G to moles of W (mol G ¡ mol W) byusing the mole ratio from the balanced equation.

3. Change the moles of W to grams of W (mol W ¡ mass W) byusing the molar mass of W.

mass G � 1 mol Gmolar mass G � mol G

mol G � b mol Wa mol G � mol W

mol W � molar mass W1 mol W � mass W

Figure 12.7 also shows the steps for doing mole-mass and mass-molestoichiometric calculations. For a mole-mass problem, the first conversion(from mass to moles) is skipped. For a mass-mole problem, the last conver-sion (from moles to mass) is skipped. You can use parts of the three-stepprocess shown in Figure 12.7 as they are appropriate to the problem youare solving.

Facts and Figures

Stoichiometry 363

DiscussInitiate a discussion with students by asking whether the law of conserva-tion of mass is always true. If not, ask them to give an example. (Nuclear reactions are the only cases where this law does not hold true.) Ask, Why isn’t there a “law of conservation of moles”? (In reactions involving rear-rangements of atoms, reactants can combine or decompose to produce fewer or greater numbers of moles of product. Although the total mass of reactants and products is constant, the number of moles of particles can increase or decrease depending on the final group-ing of atoms.) Give an example of a reaction in which the number of moles of products is greater than the number of moles of reactants.

[2H2O(l) → 2H2(g) + O2(g)]Give an example of a reaction in which the number of moles of prod-ucts is less than the number of moles of reactants.

[2Mg(s)+ O2(g) → 2MgO(s)]

Other Stoichiometric CalculationsDiscussOn the board, write equations for reac-tions in which the reactants are both gases or are a gas and a solid. Ask stu-dents how the reactants and products in each reaction would most likely be measured. Have students relate these measurements to the concept of a mole.

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Answers to...Figure 12.7 No; the given could be a product.Figure 12.8 6.02 × 1023 representa-tive particles/1 mol

Checkpoint

22.4 L

1 mol

1 mol

22.4 L

1 mol

molar mass

molar mass

1 mol


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Figure 12.8 With your knowledge of conversion factors and this problem-solving approach, you can solve a variety of stoichiometric problems. Identifying What conversion factor is used to convert moles to representative particles?

(given quantity) (wanted quantity)

aG bW

�mass

of G

1 mol Gmass G

�Volume of G

(liters) at STP

1 mol G22.4 L G

mol G mol W�b mol Wa mol G

�Representative

particles of G

1 mol G6.02�1023

�mass

of W

mass W1 mol W

�

�Volume of W

(liters) at STP

22.4 L W1 mol W

�

��Representative

particles of W

6.02�1023

1 mol W

Other Stoichiometric CalculationsAs you already know, you can obtain mole ratios from a balanced chemicalequation. From the mole ratios, you can calculate any measurement unitthat is related to the mole. The given quantity can be expressed in numbersof representative particles, units of mass, or volumes of gases at STP. Theproblems can include mass-volume, particle-mass and volume-volumecalculations. For example, you can use stoichiometry to relate volumes ofreactants and products in the reaction shown in Figure 12.8. In a typi-cal stoichiometric problem, the given quantity is first converted tomoles. Then the mole ratio from the balanced equation is used to calcu-late the number of moles of the wanted substance. Finally, the moles areconverted to any other unit of measurement related to the unit mole, asthe problem requires.

Thus far, you have learned how to use the relationship between molesand mass (1 mol � molar mass) in solving mass-mass, mass-mole, andmole-mass stoichiometric problems. The mole-mass relationship givesyou two conversion factors.

and

Recall from Chapter 10 that the mole can be related to other quantitiesas well. For example, 1 mol � 6.02 � 1023 representative particles, and1 mol of a gas � 22.4 L at STP. These two relationships provide four moreconversion factors that you can use in stoichiometric calculations.

Figure 12.8 summarizes the steps for a typical stoichiometric problem.Notice that the units of the given quantity will not necessarily be the same asthe units of the wanted quantity. For example, given the mass of G, youmight be asked to calculate the volume of W at STP.

Checkpoint What conversion factors can you write based on the mole-mass and mole-volume relationships?

1 mol molar mass

molar mass1 mol

1 mol22.4 L and 22.4 L

1 mol

6.02 � 1023 particles1 mol

1 mol6.02 � 1023 particles

and

Simulation 13 Strengthen your analytical skills by solving stoichiometric problems.

364 Chapter 12


DiscussConstruct a diagram on the board or overhead projector showing the rela-tionships that are useful for solving stoichiometry problems. One simple model reaction is A → B. Use double-headed arrows to connect the terms: Particles of A, Moles of A, Grams of A, Moles of B, Particles of B, and Grams of B. Above the appropriate arrows, write: Avogadro’s number, Coefficients, and Molar mass. Explain that the only “tran-sitions” allowed are between quanti-ties connected by arrows. Point out that the required conversion factor to make a “transition” is written above each arrow. Have students refer to the diagram when working practice problems.

Sample Problem 12.4

Answers15. 4.82 × 1022 molecules O216. 11.5 g NO2

Practice Problems PlusHydrogen gas can be made by reacting methane (CH4) with high-temperature steam:

CH4(g) + H2O(g) → CO(g) + 3H2(g)How many hydrogen molecules are produced when 158 g of methane reacts with steam? (1.78 × 1025 hydro-gen molecules)

Math HandbookFor a math refresher and practice, direct students to dimensional analysis, page R66.

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Gifted and TalentedHave students use the calculations in the sample problems as general algorithms to write computer programs that solve stoichi-ometric problems. Have students demon-strate and explain their programs to interested students.

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364 Chapter 12

Practice Problems

Practice Problems

SAMPLE PROBLEM 12.4

Calculating Molecules of a ProductHow many molecules of oxygen are produced when 29.2 g of water isdecomposed by electrolysis according to this balanced equation?


Knowns• mass of water � 29.2 g H2O• 2 mol H2O � 1 mol O2 (from balanced equation)• 1 mol H2O � 18.0 g H2O (molar mass)• 1 mol O2 � 6.02 � 1023 molecules O2

Unknown• molecules of oxygen � ? molecules O2

The following calculations need to be done:g H2O¡mol H2O¡mol O2¡molecules O2

The appropriate mole ratio relating mol O2 to mol H2O from thebalanced equation is 1 mole O2/2 mol H2O.


� 4.88 � 1023 molecules O2


The given mass of water should produce a little less than 1 mol ofoxygen, or a little less than Avogadro’s number of molecules. Theanswer should have three significant figures.

2H2O1l 2¬¡electricity2H21g 2 + O21g 2

Given Change Mole ratio Change to moleculesquantity to moles

29.2 g H2O �1 mol H2O18.0 g H2O �

1 mol O2

2 mol H2O �6.02 � 1023 molecules O2

1 mol O2

15. How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate (KClO3)?

2KClO3(s)¡2KCl(s) � 3O2(g)

16. The last step in the production of nitric acid is the reaction of nitrogen dioxide with water.

3NO2(g) � H2O(l)¡2HNO3(aq) � NO(g)

How many grams of nitrogen dioxide must react with water to produce 5.00 � 1022 moleculesof nitrogen monoxide?

Hydrogen (H2) Oxygen (O2)

The coefficients in a chemical equation indicate the relative number ofparticles and the relative number of moles of reactants and products. For areaction involving gaseous reactants or products, the coefficients also indi-cate relative amounts of each gas. As a result, you can use volume ratios inthe same way you have used mole ratios.

Math Handbook

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The electrolysis of water causes it to decompose into hydrogen and oxygen.


Stoichiometry 365

Sample Problem 12.5

Answers17. 1.93 L O218. 0.28 L PH3

Practice Problems PlusAmmonia (NH3) reacts with oxygen (O2) to produce nitrogen monoxide (NO) and water.

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)How many liters of NO are produced when 1.40 L of oxygen reacts with ammonia? (1.12 L)

CLASS ActivityCLASSCLASS ActivityCLASS

Stoichiometric Flash CardsPurpose To aid students in sequenc-ing the steps in solving stoichiometric problems

Materials 8 index cards, 1 colored index card, paper punch, 2 brass paper fasteners

Procedure Distribute the cards to the students. Have them place the eight index cards on two piles of four cards each. On the first card of the first pile, have them write Converting a given measured quantity to moles. On the each of the three remaining cards, have students write the conversion fac-tors for converting mass to moles, rep-resentative particles to moles, and volume to moles, respectively. For the second set of cards, have students label the first card Changing moles of wanted substances to a measured quan-tity. On each of the reaming cards, have them write the appropriate conversion factor. On the colored card have stu-dents write Converting moles of given to moles of wanted using mole ratio from balanced chemical equation b mol W/a mol G. Have the students use the paper pinch to punch each of the two sets of cards and attach them with a brass paper fastener. Allow students to prac-tice using the cards to solve the Prac-tice Problems.

Expected Outcome The cards should aid in sequencing the steps in solving stoichiometric problems.

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Practice ProblemsPractice Problems

SAMPLE PROBLEM 12.5

Volume-Volume Stoichiometric CalculationsNitrogen monoxide and oxygen gas combine to form the brown gasnitrogen dioxide, which contributes to photochemical smog. Howmany liters of nitrogen dioxide are produced when 34 L of oxygenreacts with an excess of nitrogen monoxide? Assume conditions of STP.

2NO(g) � O2(g) ¡ 2NO2(g)


Knowns• volume of oxygen � 34 L O2

• 2 mol NO2/1 mol O2 (mole ratio from balanced equation)• 1 mol O2 � 22.4 L O2 (at STP)• 1 mol NO2 � 22.4 L NO2 (at STP)Unknown• volume of nitrogen dioxide � ? L NO2



Because 2 mol NO2 is produced for each 1 mol O2 that reacts, thevolume of NO2 should be twice the given volume of O2. The answershould have two significant figures.

Given Change to Mole ratio Change toquantity moles liters

34 L O2 �1 mol O2

22.4 L O2�

2 mol NO2

1 mol O2�

22.4 L NO2

1 mol NO2� 68 L NO2

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17. The equation for the combus-tion of carbon monoxide is

2CO(g) � O2(g) ¡ 2CO2(g)

How many liters of oxygen are required to burn 3.86 L of car-bon monoxide?

18. Phosphorus and hydrogen can be combined to form phosphine (PH3).

P4(s) � 6H2(g) ¡ 4PH3(g)

How many liters of phos-phine are formed when 0.42 L of hydrogen reacts with phosphorus?

Did you notice that in Sample Problem 12.5 the 22.4 L/mol factorscanceled out? This will always be true in a volume-volume problem.Remember that coefficients in a balanced chemical equation indicatethe relative numbers of moles. The coefficients also indicate the relativevolumes of interacting gases.

366 Chapter 12


Sample Problem 12.6

Answers19. 18.6 mL SO2 20. 1.9 dL CO2


ASSESSEvaluate UnderstandingWrite a balanced equation on the board, such as H2(g) + I2(g) → 2HI(g). Have students write all the different mole ratios for the reaction. Then pose a problem, such as, How many moles of hydrogen iodide are formed when 0.75 mol I2 gas is reacted with excess hydrogen gas? Have students choose the correct mole ratio for the problem. Repeat with other problems and reactions, including one with gases.

ReteachUse molecular models to review the importance of mole ratios. Illustrate how the mole ratios from the balanced chemical equation are related to the individual atoms, formula units, and molecules of the reactants and prod-ucts as described by the equation.

Connecting Concepts

Figure 10.12, the “mole road map,” describes how the mass, volume, and number of representative particles of a substance can be converted into moles (and vice versa). These rela-tionships apply to six of the calcula-tions shown in Figure 12.8; the seventh is a mole-mole conversion, based on a mole ratio.

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Section 12.2 Assessment21. Mole ratios are written using the coeffi-

cients from a balanced chemical equation. They are used to relate moles of reactants and products in stoichiometric calculations.

22. Convert the given quantity to moles; use the mole ratio from the equations to find the moles of the wanted; convert moles of wanted to the desired unit.

23.

24. 176 g CO2, 36.0 g H2O

2 mol C3H7OH

9 mol O2

9 mol O2

6 mol CO2

2 mol C3H7OH

8 mol H2O

2 mol C3H7OH

6 mol CO2

6 mol CO2

2 mol C3H7OH

9 mol O2

2 mol C3H7OH

6 mol CO2

8 mol H2O

9 mol O2

8 mol H2O

8 mol H2O

6 mol CO2

8 mol H2O

9 mol O2

6 mol CO2

9 mol O2

8 mol H2O

2 mol C3H7OH

366 Chapter 12

Practice Problems

SAMPLE PROBLEM 12.6

Finding the Volume of a Gas Needed for a ReactionAssuming STP, how many milliliters of oxygen are needed to produce 20.4 mL SO3 according to this balanced equation?

2SO2(g) � O2(g)¡ 2SO3(g)


Knowns• volume of sulfur trioxide � 20.4 mL• 2 mL SO3/1 mL O2 (volume ratio from balanced equation)Unknown• volume of oxygen � ? mL O2



Because the volume ratio is 2 volumes SO3 to 1 volume O2, the volumeof O2 should be half the volume of SO3. The answer should have threesignificant figures.

Consider this equation:

CS2(l) � 3O2(g)¡ CO2(g) � 2SO2(g)

20.4 mL SO3 �1 mL O2

2 mL SO3� 10.2 mL O2

withChemASAP

19. Calculate the volume of sulfur dioxide produced when 27.9 mL O2 reacts with carbon disulfide.

20. How many deciliters of car-bon dioxide are produced when 0.38 L SO2 is formed?


21. Key Concept How are mole ratios used in chemical calculations?

22. Key Concept Outline the sequence of steps needed to solve a typical stoichiometric problem.

23. Write the 12 mole ratios that can be derived from the equation for the combustion of isopropyl alcohol.

2C3H7OH(l) � 9O2(g)¡ 6CO2(g) � 8H2O(g)

24. The combustion of acetylene gas is represented by this equation:

2C2H2(g) � 5O2(g)¡ 4CO2(g) � 2H2O(g)

How many grams of CO2 and grams of H2O are produced when 52.0 g C2H2 burns in oxygen?

Chemical Quantities Review the “mole road map“ at the end of Section 10.2. Explain how this road map ties into the summary of steps for stoichiometric problems shown in Figure 12.8.


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Stoichiometry 367

Small-ScaleLAB

Small-ScaleLAB

Analysis of Baking Soda

ObjectiveStudents calculate the mass of NaHCO3 in a sample using stoichiometry.

Prep Time 1 hour

Materials Baking soda; plastic cups; soda straws; mass balances; pipets of HCl, NaOH, and thymol blue; pH sensor (optional)

Advance Prep

Class Time 30 minutesSafety Have students wear safety glasses and follow the standard safety procedures.

Teaching Tips• Stress that the procedure measures

the amount of excess HCl that is not reacted with the baking soda (Step 4). Because this excess HCl reacts with the NaOH in a 1:1 mole ratio, the moles of NaOH equal the moles of HCl in excess. Subtracting the excess moles of HCl from the total moles used in the experiment (Step 5) yields the moles reacted with the baking soda, which is 100% NaHCO3.

• If the mixture does not turn red when thymol blue is added, the student should find the mass of a third pipet and add just enough HCl to turn the mixture cherry red. Then the student should find the mass of the half-empty pipet so the mass of HCl added can be calculated and added to the total mass used.

Expected OutcomeSample data: Step A. 2.83 g, B. 3.28 g, C. 10.70 g, D. 4.29 g, F. 10.53 g, G. 8.78 g

Solution Preparation

0.5M NaOH 20.0 g in 1.0 L

1.0M HCl 82 mL of 12M in 1.0 L Caution Always add acid to water carefully and slowly.

0.04% TB 100 mg in 21.5 mL of 01M NaOH; dilute to 250 mL

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Small-Scale Lab 367

Small-ScaleLAB

Small-ScaleLAB

PurposeTo determine the mass of sodium hydrogen carbonate in a sample of baking soda using stoichiometry.

Materials

• baking soda

• 3 plastic cups

• soda straw

• balance

• pipets of HCl, NaOH, and thymol blue

• pH sensor (optional)

Procedure

Probeware version available in the Probeware Lab Manual.

A. Measure the mass of a clean, dry plastic cup.B. Using the straw as a scoop, fill one end with baking soda

to a depth of about 1 cm. Add the sample to the cup and measure its mass again.

C. Place two HCl pipets that are about 3/4 full into a clean cup and measure the mass of the system.

D. Transfer the contents of both HCl pipets to the cup con-taining baking soda. Swirl until the fizzing stops. Wait 5–10 minutes to be sure the reaction is complete. Mea-sure the mass of the two empty HCl pipets in their cup again.

E. Add 5 drops of thymol blue to the plastic cup.F. Place two full NaOH pipets in a clean cup and measure

the mass of the system.G. Add NaOH slowly to the baking soda/HCl mixture until

the pink color just disappears. Measure the mass of the NaOH pipets in their cup again.

AnalyzeUsing your experimental data, record the answers to the following questions below your data table.

1. Write a balanced equation for the reaction between baking soda (NaHCO3) and HCl.

2. Calculate the mass in grams of the baking soda.

(Step B � Step A)

3. Calculate the total mmol of 1M HCl. Note: Every gram of HCl contains 1 mmol.

(Step C � Step D) � 1.00 mmol/g

4. Calculate the total mmol of 0.5M NaOH.Note: Every gram of NaOH contains 0.5 mmol.

(Step F � Step G) � 0.500 mmol/g

5. Calculate the mmol of HCl that reacted with the bak-ing soda. Note: The NaOH measures the amount of HCl that did not react.

(Step 3 � Step 4)

6. Calculate the mass of the baking soda from the reaction data.

(0.084 g/mmol � Step 5)

7. Calculate the percent error of the experiment.

You’re the ChemistThe following small-scale activities allow you to develop your own procedures and analyze the results.

1. Analyze It! For each calculation you did, substitute each quantity (number and unit) into the equation and cancel the units to explain why each step gives the quantity desired.

2. Design It! Baking powder consists of a mixture of bak-ing soda, sodium hydrogen carbonate, and a solid acid, usually calcium dihydrogen phosphate (Ca(H2PO4)2). Design and carry out an experiment to determine the percentage of baking soda in baking powder.

%(Step 2 - Step 62

Step 2 � 100

Analysis of Baking Soda

Analyze1. HCl + NaHCO3(s) → CO2(g) + H2O + NaCl2. 3.28 g – 2.83 g = 0.45 g3. (10.70 – 4.29)g × 1.00 mmol/g = 6.41 mmol4. (10.53 – 8.78)g × 0.500 mmol/g = 0.875

mmol (0.875 mmol HCl unreacted)5. 6.41 mmol total – 0.875 mmol unreacted =

5.53 mmol (5.53 mmol NaHCO3)6. (0.0840 g/mmol) × 5.53 mmol = 0.46 g7. (0.46 – 0.45) g/0.45g × 100% = 2% error

(assuming baking soda is one hundred per-cent sodium hydrogen carbonate)

You’re the Chemist1. See Steps 2–7.2. Repeat Steps A–G and 1–7 using baking powder instead of baking soda. The percent error is the percent of baking soda in baking powder, assuming no other errors.

For EnrichmentAsk students to predict how much baking soda and 1M HCl are needed to produce enough CO2 to fill a 1-L plastic bag. Have them write a procedure and then carry out the experiment.

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368 Chapter 12

Print• Guided Reading and Study Workbook,

Section 12.3• Core Teaching Resources, Section 12.3

Review, Interpreting Graphics• Transparencies, T133–T138• Laboratory Manual, Lab 20


Animation 13, Problem-Solving 12.25, 12.28, 12.29, 12.31, Assessment 12.3

• Go Online, Section 12.3

12.3

FOCUSObjectives12.3.1 Identify the limiting reagent in

a reaction.12.3.2 Calculate theoretical yield, per-

cent yield, and the amount of excess reagent that remains unreacted given appropriate information.

Guide for Reading

Build VocabularyLINCS Have students use the LINCS strategy for the terms theoretical yield, actual yield, and percent yield. Students should List the parts of a term they know; Imagine a picture of the term; Note a “sound-alike word” for the term; Connect the terms; and Self-test the terms.

Reading StrategyCompare and Contrast Compare and contrast the terms limiting reagent and excess reagent.

INSTRUCT

Have students study the photograph and read the text. Ask, If the carpenter decided to construct a built-in table using two legs for support, what would limit the number of tables constructed? (tabletops) What and how many unused parts would remain (3 legs)

Limiting and Excess ReagentsUse VisualsFigure 12.9 Ask, Where else do you see examples of processes or activi-ties being “limited” on an everyday basis? (Acceptable answers include the number of people you can fit into a car, and the number of cookies that can be made from a given amount of dough.)

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368 Chapter 12

12.3 Limiting Reagent and Percent Yield

If a carpenter had two table-tops and seven table legs, he would have difficulty building more than one functional four-legged table. The first table would require four of the legs, leaving just three legs for the second table. In this case, the number of table legs is the limiting factor in the construction of four-legged tables. A similar concept applies in chemistry. The amount of product made in a chemical reaction may be limited by the amount of one or more of the reactants.

Guide for Reading

Key Concepts• How is the amount of product

in a reaction affected by an insufficient quantity of any of the reactants?

• What does the percent yield of a reaction measure?

Vocabularylimiting reagent

excess reagent

theoretical yield

actual yield

percent yield

Reading StrategyBuilding Vocabulary After you have read the section, explain the differences among theoretical yield, actual yield, and percent yield.

Limiting and Excess ReagentsMany cooks follow a recipe when making a new dish. They know that suffi-cient quantities of all the ingredients must be available. Suppose, for exam-ple, that you are preparing to make lasagna and you have more thanenough meat, tomato sauce, ricotta cheese, eggs, mozzarella cheese, spin-ach, and seasoning on hand. However, you have only half a box of lasagnanoodles. The amount of lasagna you can make will be limited by thequantity of noodles you have. Thus, the noodles are the limiting ingredientin this baking venture. Figure 12.9 illustrates another example of a limitingingredient in the kitchen. A chemist often faces a similar situation. In achemical reaction, an insufficient quantity of any of the reactants will limitthe amount of product that forms.

Figure 12.9 The amount of product is determined by the quantity of the limiting reagent. In this example, the rolls are the limiting reagent. No matter how much of the other ingredients you have, with two rolls you can make only two sandwiches.

Section Resources

Stoichiometry 369


Limiting FactorPurpose Students model the concept of a limiting reagent.

Materials 15 plastic bottles, 30 plastic caps to fit the bottles, 6 containers to hold 5 caps each

Procedure Before the demonstration, place five caps in each container and put them out of the sight of the stu-dents. Set out all 15 plastic bottles. Bring out one container of caps. Ask, How many closed plastic bottles can we have? (5) What limits the number? (caps) Bring out another container of caps. Again, ask the students what will happen. Continue bringing out caps and matching them to the bottles.

Expected Outcome With two con-tainers of caps, student should indicate that the caps are the limiting factor. With three containers, students should realize that there is a one-to-one corre-spondence and therefore no limiting factor. (Or, students may argue that both the bottles and caps are equally limiting factors.) With the remaining containers, the caps are in excess and students should indicate that the bot-tles are the limiting factor.

Download a worksheet on Hydro-gen for students to complete and find additional support for NSTA SciLinks.

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Answers to...Figure 12.10 The amount would remain the same; two molecules of NH3 would form.

Checkpoint The limiting reagent is completely used up in a reaction. The excess reagent is not completely used up; some of it remains after the reaction takes place.

Less Proficient Readers Help students understand the concept of a limiting reagent by comparing it to everyday situations. For example, suppose a person is planning a dinner party. He has 6 glasses,

8 plates, and 4 chairs. How many people can he invite? Or, how many letters can you send if you have 24 pieces of stationery, 12 enve-lopes, and 10 stamps?

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Section 12.3 Limiting Reagent and Percent Yield 369

As you know, a balanced chemical equation is a chemist’s recipe. Youcan interpret the recipe on a microscopic scale (interacting particles) or ona macroscopic scale (interacting moles). The coefficients used to write thebalanced equation give both the ratio of representative particles and themole ratio. Recall the equation for the preparation of ammonia:

N2(g) � 3H2(g)¡ 2NH3(g)

When one molecule (mole) of N2 reacts with three molecules (moles) ofH2, two molecules (moles) of NH3 are produced. What would happen if twomolecules (moles) of N2 reacted with three molecules (moles) of H2? Wouldmore than two molecules (moles) of NH3 be formed? Figure 12.10 showsboth the particle and the mole interpretations of this problem.

Before the reaction takes place, nitrogen and hydrogen are present in a2:3 molecule (mole) ratio. The reaction takes place according to the bal-anced equation. One molecule (mole) of N2 reacts with three molecules(moles) of H2 to produce two molecules (moles) of NH3. At this point, all thehydrogen has been used up, and the reaction stops. One molecule (mole)of unreacted nitrogen is left in addition to the two molecules (moles) ofNH3 that have been produced by the reaction.

In this reaction, only the hydrogen is completely used up. It is thelimiting reagent, or the reagent that determines the amount of product thatcan be formed by a reaction. The reaction occurs only until the limitingreagent is used up. By contrast, the reactant that is not completely used upin a reaction is called the excess reagent. In this example, nitrogen is theexcess reagent because some nitrogen will remain unreacted.

Sometimes in stoichiometric problems, the given quantities of reac-tants are expressed in units other than moles. In such cases, the first step inthe solution is to convert each reactant to moles. Then the limiting reagentcan be identified. The amount of product formed in a reaction can bedetermined from the given amount of limiting reagent.

Checkpoint How do limiting and excess reagents differ?

Figure 12.10 The “recipe” calls for 3 molecules of H2 for every 1 molecule of N2. In this particular experiment, H2 is the limiting reagent and N2 is in excess. Inferring How would the amount of products formed change if you started with four molecules of N2 and three molecules of H2?

3H2(g)

3 molecules H2

3 mol H2

�

�

�

→

→

→

Chemical Equations

Experimental Conditions

Reactants

0 molecules H2

3 molecules H2

2NH3(g)

2 molecules NH3

2 mol NH3

Products

2 molecules NH3

0 molecules NH3

“Microscopic recipe”

“Macroscopic recipe”

After reaction

Before reaction

N2(g)

1 molecule N2

1 mol N2

1 molecule N2

2 molecules N2

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Animation 13 Apply the limiting reagent concept to the production of iron from iron ore.

For: Links on HydrogenVisit: www.SciLinks.orgWeb Code: cdn-1123

370 Chapter 12


DiscussWrite the balanced chemical equation for the following reaction on the board: 2Na(s) + Cl2(g) → 2NaCl(s).Ask a series of questions that require students to predict the amount of product formed based on molar quan-tities of reactant supplied by you. Use small whole numbers that students can easily manipulate without calcula-tors. As students solve the problems, create a list of molar quantities under each reactant and product showing how they are related for each set of conditions. In each case, ask students to state which reactant is the limiting reagent and circle the value in the list to designate it as the limiting quantity.

Sample Problem 12.7

Answers25. O2 is the limiting reagent.26. HCl is the limiting reagent.

Practice Problems PlusPhosphoric acid reacts with sodium hydroxide according to the equation:H3PO4(aq) + 3NaOH(aq) →

Na3PO4(aq) + 3H2O(l)If 1.75 mol H3PO4 is made to react with 5.00 mol NaOH, identify the limiting reagent. (5.25 mol NaOH required; NaOH is the limiting reagent)


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370 Chapter 12

Practice Problems

Practice Problems

SAMPLE PROBLEM 12.7

Determining the Limiting Reagent in a ReactionCopper reacts with sulfur to form copper(I) sulfide according to thefollowing balanced equation.

2Cu(s) � S(s)¡ Cu2S(s)

What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?

Analyze List the knowns and the unknown.Knowns• mass of copper � 80.0 g Cu• mass of sulfur � 25.0 g S

Unknown• limiting reagent � ?

The number of moles of each reactant must first be found:

g Cu ¡mol Cu

g S ¡mol S

The balanced equation is used to calculate the number of moles of onereactant needed to react with the given amount of the other reactant:

mol Cu ¡mol S

The mole ratio relating mol S to mol Cu from the balanced chemicalequation is 1 mol S/2 mol Cu.


Comparing the amount of sulfur needed (0.630 mol S) with the givenamount (0.779 mol S) indicates that sulfur is in excess. Thus copper isthe limiting reagent.

Evaluate Do the results make sense?

Since the ratio of the given mol Cu to mol S was less than the ratio (2:1)from the balanced equation, copper should be the limiting reagent.

80.0 g Cu � 1 mol Cu63.5 g Cu � 1.26 mol Cu

25.0 g S � 1 mol S32.1 g S � 0.779 mol S

Given Mole Neededquantity ratio amount

1.26 mol Cu � 1 mol S2 mol Cu � 0.630 mol S

25. The equation for the com-plete combustion of ethene (C2H4) is

C2H4(g) � 3O2(g)¡2CO2(g) � 2H2O(g)

If 2.70 mol C2H4 is reacted with 6.30 mol O2, identify the limiting reagent.

26. Hydrogen gas can be pro-duced by the reaction of magnesium metal with hydrochloric acid.

Mg(s) � 2HCl(aq)¡MgCl2(aq) � H2(g)

Identify the limiting reagent when 6.00 g HCl reacts with 5.00 g Mg.

Sulfur (S) Copper (Cu)

Copper(I) sulfide (Cu2S)

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Math Handbook


Stoichiometry 371

RelateHave students reread the Inquiry Activ-ity. Ask, What is the maximum num-ber of product molecules that can be formed using the procedure described in the activity? (Given that (a) one molecule of M2 reacts with three molecules of C2, (b) there are only ten molecules of each reactant available, and (c) fifteen molecules are chosen, a maximum of three M2 molecules can react with nine C2 molecules to form a total of six MC3 molecules.)

Sample Problem 12.8

Answers27. a. 5.40 mol O2 required; C2H4 is the

limiting reactant.b. 5.40 mol H2O

28. 43.2 g H2O


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Comparative Torch Temperatures• Acetylene welding torches produce flame

temperature of 3300–3400°C.• Propane torches reach flame temperatures

of about 1400°C.• The flame of a match has a temperature

range of 600–800°C.

Facts and Figures


Practice Problems

Practice Problems

27. The equation below shows the incomplete combustion of ethene.

C2H4(g) � 2O2(g)¡2CO(g) � 2H2O(g)

If 2.70 mol C2H4 is reacted with 6.30 mol O2,a. identify the limiting reagent.b. calculate the moles of water

produced.

28. The heat from an acetylene torch is produced by burning acetylene (C2H2) in oxygen.

2C2H2(g) � 5O2(g)¡4CO2(g) � 2H2O(g)

How many grams of water can be produced by the reaction of 2.40 mol C2H2 with 7.4 mol O2?

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SAMPLE PROBLEM 12.8

Using a Limiting Reagent to Find the Quantity of a ProductWhat is the maximum number of grams of Cu2S that can be formedwhen 80.0 g Cu reacts with 25.0 g S?

2Cu(s) � S(s)¡ Cu2S(s)


Knowns• limiting reagent � 1.26 mol Cu (from Sample Problem 12.7)• 1 mol Cu2S � 159.1 g Cu2S (molar mass)Unknown• mass copper(I) sulfide � ? g Cu2S

The limiting reagent, which was determined in the previous sampleproblem, is used to calculate the maximum amount of Cu2S formed:

mol Cu ¡mol Cu2S¡ g Cu2S

The equation yields the appropriate mole ratio: 1 mol Cu2S/2 mol Cu.


The given quantity of copper, 80.0 g, could have been used for this stepinstead of the moles of copper, which were calculated in SampleProblem 12.7.

Evaluate Do the results make sense?

Copper is the limiting reagent in this reaction. The maximum numberof grams of Cu2S produced should be more than the amount of copperthat initially reacted because copper is combining with sulfur. How-ever, the mass of Cu2S produced should be less than the total mass ofthe reactants (105.0 g) because sulfur was in excess.

1.26 mol Cu �1 mol Cu2S2 mol Cu �

159.1 g Cu2S1 mol Cu2S � 1.00 � 102 g Cu2S


In Sample Problem 12.7, you may have noticed that even though themass of copper used in the reaction is greater than the mass of sulfur, cop-per is the limiting reagent. The reactant that is present in the smalleramount by mass or volume is not necessarily the limiting reagent.

372 Chapter 12


Quick LABQuick LAB

Limiting ReagentsObjective After completing this activ-ity, students will be able to• compare the amount of hydrogen gas

produced from the reaction of mag-nesium and hydrochloric acid with the amounts of magnesium.

• infer the limiting reagent in reactions of magnesium and hydrochloric acid.

Skills Focus Observing, measuring, calculating

Prep Time 15 minutes

Materials graduated cylinders, mass balances, 250-mL Erlenmeyer flasks, rubber balloons, magnesium ribbons, 1.0M hydrochloric acid

Class Time 20 minutes

Safety The balloons contain hydro-gen gas and should be kept away from heat and open flames during and after the experiment. Caution students about the corrosive nature of HCl. Stu-dents should wear safety glasses and aprons. Once the reaction in a flask is complete, carefully remove the bal-loon. Disperse the gas in a fume hood or outside. Neutralize remaining HCl(aq) with baking soda before flush-ing down the drain.

Teaching Tips• Have students carefully balance the

equation for the reaction and then look at the mole ratio from the bal-anced equation and the molar amounts of the two reactants present before they make their prediction in Step 4.

Expected Outcome The approximate volumes of the balloons are 0.60 g Mg, 0.5 L H2; 1.2 g Mg, 1 L H2; 2.4 g Mg, 1 L H2.

Analyze and Conclude 1. Those students who failed to bal-

ance the equation, obtain the mole ratio, and compare the molar amounts of reactants, likely pre-dicted a doubling of the volume with each doubling of the mass of magnesium.

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2. Mg + 2HCl → MgCl2 + H23. A mass of 0.60 g Mg is 0.025 mol Mg. Because

Mg and HCl react in a 1:2 mol ratio, 0.10 mol HCl is in excess and Mg is limiting. According to the balanced equation, 0.025 mol Mg should produce 0.025 mol H2. A mass of 1.2 g Mg is 0.050 mol Mg. According to the bal-anced equation, 0.050 mol Mg will react with 0.1 mol HCl to produce 0.05 mol H2.

4. A mass of 2.4 g Mg is 0.10 mol Mg. Because there is only 0.1 mol HCl in the flask and, according to the equation, 0.20 mol HCl is needed to react with 0.10 mol Mg, HCl is limit-ing. Therefore, only 0.05 mol H2 is produced.

For EnrichmentAsk, What other clue could be used to deter-mine if HCl was the limiting reagent? (If HCl were the limiting reagent, remaining magnesium metal would be present after the reaction.)

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372 Chapter 12

Quick LABQuick LAB

Procedure1. Add 100 mL of the hydrochloric acid

solution to each flask.

2. Weigh out 0.6 g, 1.2 g, and 2.4 g of magnesium ribbon, and place each sample into its own balloon.

3. Stretch the end of each balloon over the mouth of each flask. Do not allow the magnesium ribbon in the balloon to fall into the flask.

4. Magnesium reacts with hydrochloric acid to form hydrogen gas. When you mix the magnesium with the hydro-chloric acid in the next step, you will generate a certain volume of hydrogen gas. How do you think the volume of hydrogen produced in each flask will compare?

5. Lift up on each balloon and shake the magnesium into each flask. Observe the volume of gas produced until the reaction in each flask is completed.

Limiting Reagents

PurposeTo illustrate the concept of a limiting reagent in a chemical reaction.

Materials

• graduated cylinder

• balance

• 3 250-mL Erlenmeyer flasks

• 3 rubber balloons

• 4.2 g magnesium ribbon

• 300 mL 1.0M hydrochlo-ric acid

Analysis and Conclusions1. How did the volumes of hydrogen gas

produced, as measured by the size of the balloons, compare? Did the results agree with your prediction?

2. Write a balanced equation for the re-action you observed.

3. The 100 mL of hydrochloric acid con-tained 0.10 mol HCl. Show by calcula-tion why the balloon with 1.2 g Mg inflated to about twice the size of the balloon with 0.60 g Mg.

4. Show by calculation why the balloons with 1.2 g and 2.4 g Mg inflated to approximately the same volume. What was the limiting reagent when 2.4 g Mg was added to the acid?

Percent YieldIn theory, when a teacher gives an exam to the class, every student shouldget a grade of 100%. This generally does not occur, as shown in Figure 12.11.Instead, the performance of the class is usually spread over a range ofgrades. Your exam grade, expressed as a percentage, is a ratio of two items.The first item is the number of questions you answered correctly. The sec-ond is the total number of questions. The grade compares how well youperformed with how well you could have performed if you had answered allthe questions correctly. Chemists perform similar calculations in the labo-ratory when the product from a chemical reaction is less than expected,based on the balanced chemical equation.

When an equation is used to calculate the amount of product that willform during a reaction, the calculated value represents the theoreticalyield. The theoretical yield is the maximum amount of product that couldbe formed from given amounts of reactants. In contrast, the amount ofproduct that actually forms when the reaction is carried out in the labora-tory is called the actual yield. The percent yield is the ratio of the actualyield to the theoretical yield expressed as a percent.

Percent yield �actual yield

theoretical yield � 100%

Figure 12.11 Calculating the ratio of the number of correct answers to the number of questions on the exam is a measure of how well the student performed on the exam.

Stoichiometry 373

Percent Yield

CLASS ActivityCLASS

Actual Yield and HeatPurpose Students assess actual yield of a reaction based on temperature changes

Materials 3 styrofoam cups, a ther-mometer, 100 mL of 1.0M HCl, and approximately 200 mL of 1.0M NaOH

Safety Students should wear safety glasses and aprons. HCl and NaOH are corrosive. Caution students to avoid skin contact with these chemicals. Neutralized solutions can be flushed down the drain with excess water.

Procedure Write the balanced chemi-cal equation on the board for students to use as a reference: HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) + heat. Have students interpret the balanced equation as 1 mole of HCl reacts with 1 mole of NaOH to form 1 mole of NaCl and 1 mole of water. Have students transfer 30 mL of the HCl to a styro-foam cup and measure the tempera-ture. Next, have them add 10 mL of NaOH and gently stir the contents with the thermometer. Students should record the highest temperature reached as the reaction proceeds. Using a new cup each time, repeat the measurement with 30 mL of HCl and 30 mL of NaOH, and with 30 mL of HCl and 60 mL of NaOH.Expected Outcomes The second trial, in which reactants are present in a 1:1 mole ratio, produces the greatest tem-perature change. A slightly lower tem-perature change for the third trial can be accounted for by the greater mass of reagents. Ask, Which trial had the highest yield? (the one that produced the greatest temperature change)

Answers to...Figure 12.13 H2O and CO2

Checkpoint

Reactions do not always go to completion; side reactions may occur; poor labora-tory technique.


Because the actual yield of a chemical reaction is often less than thetheoretical yield, the percent yield is often less than 100%. The percentyield is a measure of the efficiency of a reaction carried out in the laboratory.This is similar to an exam score measuring your efficiency of learning, or abatting average measuring your efficiency of hitting a baseball.

A percent yield should not normally be larger than 100%. Many factorscause percent yields to be less than 100%. Reactions do not always go tocompletion; when this occurs, less than the calculated amount of productis formed. Impure reactants and competing side reactions may causeunwanted products to form. Actual yield can also be lower than the theo-retical yield due to a loss of product during filtration or in transferringbetween containers. Moreover, if reactants or products have not been care-fully measured, a percent yield of 100% is unlikely.

An actual yield is an experimental value. Figure 12.13 shows a typicallaboratory procedure for determining the actual yield of a product of adecomposition reaction. For reactions in which percent yields have beendetermined, you can calculate and therefore predict an actual yield if thereaction conditions remain the same.

Checkpoint What factors can cause the actual yield to be less than the theoretical yield?

Figure 12.12 A batting average is actually a percent yield.

a b c

Figure 12.13 Sodium hydrogen carbonate (NaHCO3) will decompose when heated. The mass of NaHCO3, the reactant, is measured. The reactant is heated. The mass of one of the products, sodium carbonate (Na2CO3), the actual yield, is measured. The percent yield is calculated once the actual yield is determined. Predicting Whatare the other products of this reaction?

a

bc

374 Chapter 12


374 Chapter 12

Practice Problems

Practice Problems

Recall that the percent yield is calculated by multiplying the ratio of theactual yield to theoretical yield by 100%. Therefore, you must have valuesof both the theoretical yield and the actual yield to calculate the percentyield.

SAMPLE PROBLEM 12.9

Calculating the Theoretical Yield of a ReactionCalcium carbonate, which is found in seashells, is decomposed byheating. The balanced equation for this reaction is:

What is the theoretical yield of CaO if 24.8 g CaCO3 is heated?


Knowns• mass of calcium carbonate � 24.8 g CaCO3

• 1 mol CaCO3 � 100.1 g CaCO3 (molar mass)• 1 mol CaO � 56.1 g CaO (molar mass)Unknown• theoretical yield of calcium oxide � ? g CaOCalculate the theoretical yield using the mass of the reactant:

g CaCO3¡mol CaCO3¡mol CaO ¡ g CaO

The appropriate mole ratio is 1 mol CaO/1 mol CaCO3.



The mole ratio of CaO to CaCO3 is 1:1. The ratio of their masses in thereaction should be the same as the ratio of their molar masses, whichis slightly greater than 1:2. The result of the calculations shows that themass of CaO is slightly greater than half the mass of CaCO3.

CaCO3(s 2D¡ CaO(s 2 + CO2(g 2

� 13.9 g CaO

24.8 g CaCO3 �1 mol CaCO3

100.1 g CaCO3� 1 mol CaO

1 mol CaCO3�

56.1 g CaO1 mol CaO

29. When 84.8 g of iron(III) oxide reacts with an excess of carbon monoxide, iron is produced

Fe2O3(s) � 3CO(g)¡ 2Fe(s) � 3CO2(g)

What is the theoretical yield of this reaction?

30. When 5.00 g of copper reacts with excess silver nitrate, sil-ver metal and copper (II) nitrate are produced. What is the theoretical yield of silver in this reaction?

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Math Handbook



DiscussPoint out the importance of the yield in a chemical reaction. Note that actual yields are usually less than theoretical yields. For industrial chemists or chem-ical engineers, the goal is to find cost-effective methods for converting reac-tants into products. Industrial chemists and chemical engineers want to achieve the maximum product yield at the lowest cost. One way to control costs is to minimize the amount of excess reagent by calculating stoichio-metric quantities of the reactants.

Sample Problem 12.9

Answers29. 59.3 g Fe30. 17.0 g Ag



Sample Problem 12.10

Answers31. 83.5%32. 57.7%

Practice Problems PlusIf 75.0 g of siderite ore (FeCO3) is heated with an excess of oxygen, 45.0 g of ferric oxide (Fe2O3) is pro-duced.4FeCO3(s) + O2(g) →

2Fe2O3(s) + 4CO2(g) What is the percent yield of this reaction? (87.0%)

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Stoichiometry 375

ASSESSEvaluate Understanding

Limiting ReactionFollowing the safety precaution listed in the Quick Lab on page 372, place a small piece of zinc in a large beaker of dilute hydrochloric acid. Explain the reaction to the students.

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

Ask, Which of the two reactants is the limiting reagent? (If the zinc com-pletely disappears after some time, stu-dents may conclude that the zinc is used up first, and is, therefore, the limiting reagent.) After students respond, ask, How can you test your hypothesis? (Acceptable answers include measure the volume of H2 produced for various amounts of zinc added to a fixed volume of HCl. Determine the stoichiometric quantity of zinc, the amount that reacts to give the greatest volume of H2.)

ReteachEmphasize the importance of having a correctly balanced equation in order to properly calculate the theoretical yield for a reaction. Explain that the way to decide which substance, if any, is the lim-iting reagent, is to compare the mole ratios of the given substances to the required mole ratio, as shown in the bal-anced chemical equation.

Elements Handbook

the amounts of N2 and H2 used (to calculate the theoretical yield), and the amount of NH3 produced (i.e., the actual yield)

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Section 12.3 Assessment33. In a chemical reaction, an insufficient

quantity of any of the reactants will limit the amount of product that forms.

34. The efficiency of a reaction carried out in a laboratory can be measured by calcu-lating the percent yield.

35. 70.5%


Handbook

Practice Problems

Practice Problems

SAMPLE PROBLEM 12.10

Calculating the Percent Yield of a ReactionWhat is the percent yield if 13.1 g CaO is actually produced when 24.8 gCaCO3 is heated?


Knowns• actual yield � 13.1 g CaO• theoretical yield � 13.9 g CaO (from Sample Problem 12.9)

•

Unknown• percent yield � ? %



In this example, the actual yield is slightly less than theoretical yield.Therefore, the percent yield should be slightly less than 100%. Theanswer should have three significant figures.

CaCO3(s 2D¡ CaO(s 2 + CO2(g 2

percent yield �actual yield

theoretical yield � 100%

percent yield �13.1 g CaO13.9 g CaO � 100% � 94.2%

31. If 50.0 g of silicon dioxide is heated with an excess of car-bon, 27.9 g of silicon carbide is produced.

SiO2(s) � 3C(s)¡ SiC(s) � 2CO(g)What is the percent yield of this reaction?

32. If 15.0 g of nitrogen reacts with 15.0 g of hydrogen, 10.5 g of ammonia is produced. What is the percent yield of this reaction?


33. Key Concept In a chemical reaction, how does an insufficient quantity of a reactant affect the amount of product formed?

34. Key Concept How can you gauge the effi-ciency of a reaction carried out in the laboratory?

35. What is the percent yield if 4.65 g of copper is produced when 1.87 g of aluminum reacts with an excess of copper(II) sulfate?

2Al(s) � 3CuSO4(aq)¡ Al2(SO4)3(aq) � 3Cu(s)

Haber Process Read the feature on ammonia on page R26. Examine the flow chart summarizing the Haber process. What experimental data would you need to determine the percent yield of the Haber process?

withChemASAP

withChemASAP



Math Handbook

For help with percents, go to page R72.

376 Chapter 12

Just the Right Volume of GasWrite the following reactions on the board.

Reaction I 2NaN3(s) → 2Na(s) + 3N2(g) Reaction II 10Na(s) + 2KNO3(s) →

K2O(s) + 5Na2O(s) + N2(g) + heat

Point out that the proper inflation of the air bag requires two reactions. Explain that an electrical current pro-duced by the igniter causes the decomposition of sodium azide into sodium metal and nitrogen gas. Note that the sodium metal produced is dangerously reactive. In a second reac-tion, potassium nitrate reacts with the elemental sodium and forms potas-sium oxide, sodium oxide, and addi-tional nitrogen gas. The heat causes all the solid products to fuse with SiO2, powdered sand, which is also part of the reaction mixture. The fused prod-uct is a safe, unreactive glass.

DiscussAsk, How many moles of potassium nitrate must be included in the reac-tion mixture to consume the sodium produced by the decomposition of one mole of sodium azide? (0.2 mol KNO3) How many liters of N2 are pro-duced at STP if 1.0 mole of sodium azide and 0.20 mole of potassium nitrate react? (36 L) Have students speculate how the pressure of the gas inside the air bag depends on the number of moles of nitrogen produced and the temperature inside the air bag. (Acceptable answers should indicate that because gas pressure depends on the number of gas particles present, pressure depends on the number of moles of gas particles present. The heat released by this reaction raises the tem-perature of the gaseous products, help-ing the bag inflate even faster.)

L2

Automobile Restraint Systems

History

1947 Tucker Automobile Company makes safety belts available.

1949 Nash Motor Company provides lap-safety belts.

1958 Swedish engineer, Nils Bohlin, patents chest/lap-safety belt.

1963 Volvo makes chest/lap-safety belts standard equipment on its U.S. models.

1988 Chrysler Motor Company makes air-bag restraint systems standard equipment.

Facts and Figures

376 Chapter 12

Just the Right Volume of Gas

In a front-end collision, proper inflation of an air bag may

save your life. Engineers use stoichiometry to determine

the exact quantity of each reactant in the air bag's inflation

system. Interpreting Diagrams What is the source of the

gas that fills an air bag?

Car Facts

17.1 million cars and light trucks were sold in 2002 in the United States.

In the United States, the Cadillac Escalade SUV has the most frequent theft claims among 2000–02 model passenger vehicles.

Monaco has the highest number of vehicles in relation to its road network. In 1996 (most recent figures), it had 480 vehicles for each kilometer of road. If they were required to park behind one another on the streets, half would have nowhere to park!

At night, headlights illuminate 160 ft in front of your car. If you are driving 40 mi/h at night, your reaction distance is 88 ft and your braking distance is 101 ft. Is your stopping distance less than or greater than 160 ft?

Reactiondistance

�Brakingdistance

�Stoppingdistance

Answers to...

Interpreting DiagramsThe sources of the N2 gas are the decomposition reaction of sodium azide (NaN3) and the reaction of sodium with potassium nitrate.

Stoichiometry 377

DiscussHave students research the safety restrictions on the use of air bags with infants and young children. Have stu-dents research new designs and appli-cations of safety air bags.

L2

Automobile Restraint SystemsStatistic The use of air-bag restraint sys-tems (seatbelt/air bags) reduces the risk of fatalities in accidents by about 70%.

Facts and Figures

Steeringwheel

Air bag folded intosteering wheel

Ignition unit

Igniter

Sodium azidepellets

Nitrogen gas

Steering wheel

Steering wheel

Electrical signalfrom crash sensor

Igniter

Sodium azide pelletsdecomposing2NaN3¡ 2Na � 3N2

Deflated air bag

Nontoxic residue ofsecondary reactions

Technology and Society 377

A collision triggers crash sensors, which send a signal

to an igniter.

The igniter triggers a series of chemical reactions that

release a large volume of nitrogen gas, which fills the air bag. Within 0.05 seconds of the collision the air bag is fully inflated.

Small holes in the bag allow nitrogen gas to

escape, causing the bag to deflate.

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