Chapter 8 Quantities In Reactions. Homework Assigned Problems (odd numbers only) “Problems” 17...

Chapter 8Quantities

In Reactions

Homework• Assigned Problems (odd numbers only)

• “Problems” 17 to 73• “Cumulative Problems” 75-95• “Highlight Problems” 97-99

Calculations Using Balanced Equations: Stoichiometry

• Stoichiometry is the study of the quantitative relationships among reactants and products in a chemical reaction

• These chemical calculations can be used to determine the amount of one reactant needed to completely react with another

• Or, to determine the amount of reactant needed to produce a desired amount of product

• To calculate chemical quantities in reactions involves knowing how to interpret a balanced chemical equation

Calculations Using Balanced Equations: Law of Conservation of Mass

• As a chemical reaction proceeds:–Reactants are consumed and new

materials with new chemical properties are produced

–Bonds are broken, formed, or atoms are rearranged which produces new substances

–No material is lost or gained as original substances change to new substances

Calculations Using Balanced Equations: Law of Conservation of Mass

• Law of Conservation of Mass–Quantity of matter does not change

during a chemical reaction–The sum of the masses of products is

equal to the sum of masses of reactants

–Atoms are neither created nor destroyed in chemical reactions

–Only a balanced equation obeys this law

Mole Relationships in Chemical Equations: Conservation of Mass

• A balanced equation has the same number of atoms on each side of the arrow

(g)OH (g)CO (g)O (g)CH 2224 22

Information Available from a Balanced Chemical Equation

• “1 mole of methane gas reacts with 2 moles of oxygen gas to produce 1 mole of carbon dioxide and 2 moles of water vapor.”

• Multiplying each of the molar masses by the coefficient will give the total mass of reactants and products

Making Molecules:Mole-Mole Conversions

A balanced chemical equations tell us:– The formulas and symbols of the reactants and

products– The physical state of each substance– If special conditions such as heat are required– The number of molecules, formula units, or atoms

of each type of molecule involved in the reaction • Number can be in terms of single atoms, or

moles of atoms– The relative number of moles of each reactant and

product

Making Molecules:Mole-Mole Conversions

• In a balanced equation, conversion from moles of one substance to another will be determined by the values of the coefficients

• Balancing an equation will generate the coefficients that equal the number of moles of each reactant and product

• To determine how many moles of methanol would be produced if 0.295 moles of hydrogen gas is consumed:

• Requires a mole to mole relationship between methanol and hydrogen gas

)OH(CH (g)H CO(g) 32 2

Making Molecules:Mole-Mole Factors

• Converting the given mole amount of hydrogen gas enables you to find the number of moles methanol produced

• Coefficients of the balanced equation can be used to make mole to mole relationships between the different reactants and products

)((g)(g) OHCH H CO 32 2

Given: 0.295 mol H2 Find: mol CH3OH

2 mol H2 = 1 mol CH3OH


• The unit path begins with moles of H2 and ends with moles of CH3OH

• From any mole-mole relationship, two mole-mole conversion factors can be made:

)((g)(g) OHCH H CO 32 2

mol H2 mol CH3OH

2

3

3

2

H mol 2

OHCH mol 1

OHCH mol 1

H mol 2or

2

3

H mol 2

OHCH mol 1

Mole-molefactor

2 mol H2 = 1 mol CH3OH


• Write all of the possible mole-mole factors for the following chemical equation

)((g)(g) OH O H 222 2 2

2 mol H2

1 mol O2

2 mol H 2

2 mol H2O1 mol O2

2 mol H2O

1 mol O2

2 mol H2

2 mol H2O2 mol H 2 2

2

O mol 1

OH mol 2

2 mol H2 = 2 mol H2O 1 mol O2 = 2 mol H2O2 mol H2 = 1 mol O2

Using Mole-Mole Factors in Calculations:Calculating Moles of a Product

• Calculations based on balanced equations require the use of mole to mole (conversion) factors– Equation must be balanced

– Identify the known (given) and needed (find) substances

– Make the conversion factor based on:

KNOWN is whatof tcoefficien

SOUGHTis whatof tcoefficienmolesofnumbergiven )(KNOWN


• Using mole-mole factors from a BALANCED chemical equation–You can convert moles of one

compound to moles of another compound using the correct mole-mole factor

moles A moles B

grams A grams B

MM of A MM of B

moles B moles A

Stoichiometry

Mole-molefactor


• Calculate the moles of CO2 formed when 4.30 moles of C3H8 reacts with (the required) 21.5 moles of O2

• Balance the equation• Plan to convert the given amount of moles to the

needed amount of moles• Use coefficients to state the relationships and

mole-mole factors• Set up the problem using the mole-mole factor

and canceling units

O(g)H (g)CO (g)O (g)HC 22283 5 3 4


83

2

2

83

HC mol 1

CO mol 3

CO mol 3

HC mol 1and

moles A moles B

grams A grams B

MW of A MW of B

moles A moles B

Stoichiometry

83HC mol 4.302CO mol

C3H8(g) O2 (g) CO2 (g) H2O(g)5 3 4

Given: Find:

1 mol C3H8 = 3 mol CO2

Mole-molefactor

Mole-mole factorMole-mole relation

MM of A MM of B

4.30 mol C3H8 3 mol CO 2

1 mol C3H8


• Set up the problem using the mole-mole factor that cancels given moles and provides needed moles

C3H8(g) O2 (g) CO2 (g) H2O(g)5 3 4

83HC mol 4.30 :given 2CO mol :find

12.9 mol CO 2

Mole-mole factor83

2

HC mol 1

CO mol 3

Making Molecules: Mass-to-Mass Conversions

• From the balanced equation–It is also possible to start with a known

mass of one substance–Then convert to moles of another

substance–Start with a given amount (of grams) of

a substance–Then use mole-mole factor to find the

sought-after mass of another substance

Making Molecules: Mass-to-Mass Conversions

• The most common type of stoichiometric calculation is the mass-to-mass calculation

• In this type of problem: • The mass of one substance involved in a

chemical reaction is given• Find the mass of another substance involved in

the reaction• If a chemist only has so many grams of a certain

chemical – Can calculate how many grams of another substance can

be produced– Can calculate how many grams of another reactant are

required to react with it

Mass-to-Mass Conversions

• To convert the grams of one substance to grams of another substance:

• Find the mole-mole factor using the coefficients in the balanced equation– You can only relate (moles-moles) of two compounds, not

grams-to-grams– Ratios ONLY apply to moles, NOT grams– Must convert grams to moles, then use mole-mole factor

moles A moles B

grams A grams B

MW of A MW of B

moles A moles B

StoichiometryMM of A MM of B

moles A

moles BMole-molefactor

Mass-to-Mass Conversions

• Mass-to-mass conversions begin with a given mass of substance A

• By use of the balanced equation, find the mass of another (substance B)1. Convert grams of A to moles of A2. Convert the moles of A to moles of B by use of mole-mole

ratio3. Convert moles of B to mass of B

moles A moles B

grams A grams B

MM of A MM of B

moles B moles A

Stoichiometry

Mass-to-Mass Conversions(Mass of Product from Mass of Reactant)

• Calculate the mass of carbon dioxide produced when 96.1 g of propane react with sufficient oxygen.

• Balance the equation• Plan to convert the given mass to given moles• Convert the given moles to sought-after moles by the use of mole-

mole factor

• Convert the needed moles to needed mass

C3H8(g) O2 (g) CO2 (g) H2O(g)5 3 4

moles C3H8moles CO2

grams C3H8 grams CO2

MM of C3H8

MM of CO2

moles CO2

moles C3H8

Stoichiometry

Mole-molefactor

Mass-to-mass Conversions: Example 1• Write the equalities

• 1 mol C3H8 = 44.09 g C3H8

• 1 mol CO2 = 44.01 g CO2

• 1 mol C3H8 = 3 mol CO2 to create mole-mole factor

C3H8(g) O2 (g) CO2 (g) H2O(g)5 3 4Find: g of CO2Given: 96.1 g C3H8

moles C3H8moles CO2

grams C3H8 grams CO2

MM of A MM of B

moles B moles A

Stoichiometry

83

2

HC mol 1

CO mol 3

Mass-to-Mass Conversions: Example 1

C3H8(g) O2 (g) CO2 (g) H2O(g)5 3 496.1 g C3H8

2.1796 mol C 3H8

6.539 mol CO2

2CO g 288

83

8383

HC g 44.09

HC mol 1HC g 96.1

2.1796 mol C3H8 3 mol CO2

1 mol C3H8

6.539 mol CO2 44.01 g CO2

1 mol CO2

X g CO2


• What mass of carbon monoxide and what mass of hydrogen are required to form 6.0 kg of methanol by the following reaction:

)OH(CH (g)H CO(g) 32 2

CO g )(x 2H g )( y 6.0 kg CH3OH


)(OHCH (g)H (g)CO 32 l 2

6000 g CH3OH

187.27 mol CH3OH

187.27 mol CO

374.53 mol H2

OHCH kg 1

g 1000 OHCH kg 6.0

3

3

6000 g CH3OH 1 mol CH 3OH

32.04 g CH3OH

187.27 mol CH3OH 1 mol CO1 mol CH3OH

187.27 mol CH 3OH 2 mol H 2

1 mol CH 3OH

Mass Calculations Example 2

CO g 5200

2H g 750

187.27 mol CO 28.01 g CO1 mol CO

2

22

H mol 1

H g 2.016 H mol 374.53

Limiting Reactant, Theoretical Yield, and Percent Yield

• The chemical reactants are usually not present in the exact mole-mole ratios as stated in the balanced chemical equation

• Often, one of the reactants is purposely added in an excess amount

• Reasons include:• Increase the rate of reaction• To ensure that one reactant is completely used

up (reacted)• Reactants are not completely converted to

products as stated on paper (theory)

Limiting Reactant, Theoretical Yield, and Percent Yield

• Chemical reactions with two or more reactants will continue until one of the reactants is used up (consumed)

• If one of the reactants is used up, the reaction will stop because there is not enough of the other reactant to react with it

• The reactant used up is called the limiting reactant (reagent)

• This reactant limits the amount of product that can be made

Limiting Reactant

• When you make peanut butter sandwiches

• Required: 2 slices of bread and 1 tbsp. peanut butter per sandwich

• The reaction of nitrogen gas and hydrogen gas forming ammonia gas

• Required: 1 molec. N2 gas and 3 molec. H2 gas

2 slices of br. +1 tbsp p.b. =1sndw.

N2 (g) + 3H2 (g) = 2 NH3 (g)

N2

NH3

NH3

Limiting Reactant

To determine the limiting reactantbetween two reactants:

1. Balance the equation 2. State the mole-mole relationships to

make conversion factors3. Convert the initial masses (reactants)

to moles of each reactant4. Calculate how many moles of product

can be produced by each reactant

Limiting Reactant

5. Convert the moles of product to number of grams of product that each of the reactants would produce

6. Compare the numbers: The reactant producing the least amount of product (grams) is the limiting reactant

Limiting Reactant Problem• Lithium nitride, an ionic compound

containing Li+ and N3- ions, is prepared by the reaction of lithium metal and nitrogen gas. Calculate the mass of lithium nitride formed from 56.0 g of nitrogen gas and 56.0 g of lithium metal.

(s)(g)(s) NLi N Li 32 2656.0 g Li 56.0 g N2

NLi g 3X

Limiting Reactant Problem

6 2Find: g of Li3NGiven: 56.0 g Li Given: 56.0 g N2

moles Li moles Li3N

grams Li grams Li3N

MM of Li MM of Li3N

moles Li3N moles Li

Stoichiometry

1 mol Li = 6.941g Li 1 mol N2 = 28.00 g N2

Li(s) N2(g) Li3N(s)

1 mol Li3N= 34.83 g Li3N

Equalities and Conversion Factors-

Solution Map:

MM of N2

moles N2

moles N2

grams N2

6 mol Li = 2 mol Li3N 1 mol N2 = 2 mol Li3N


NLimol

NLig

Limol

NLimolLimol

3

33

1

82.34

6

207.8

NLi g (x) N g 56.0 Li g 56.0

N(s)2Li(g)N (s) Li6

32

32

56.0 g Li 1 mol Li6.941 g Li

56.0 g N2 1 mol N 2

28.02 g N2

93.67 g Li3N

139.3 g Li3N

Limiting reactant

8.07 mol Li

2.00 mol N2

NLimol

NLig

Nmol

NLimolNmol

3

3

2

32 1

82.34

1

200.2


• Lithium is the limiting reactant. We calculated the number of grams of lithium nitride which is formed in the reaction based on the limiting reactant

• This is the calculated amount of lithium nitride formed if the reaction proceeds completely as described by its balanced chemical equation

93.7 g Li3NNLimol

NLigNLimol

3

33 1

83.3469.2

Theoretical yield

Percent Yield• The calculated amount of product that

should be obtained is called the theoretical yield

• Assumes all reactants are converted to product based on the mole-mole ratios of reactant to product

• Rarely do you get the maximum amount of product– Side reactions– Loss during transfer– Accidental spills

Percent Yield• Theoretical Yield

– The calculated amount of product• Actual Yield

– The actual amount of product– Something less than the theoretical

• Percent Yield– The fraction of the theoretical yield actually

obtained is expressed as a percent

% Yield Actual Yield

Theoretical Yield100%

Percent Yield Example• Suppose, in the previous limiting

reactant problem, you actually produced 90.8 g of Li3N. What is the percent yield of this reaction?

%1007.93

8.90

g

g96.9 % yield %100

ltheoretica

actual

Enthalpy• Chemical reactions are associated with an

absorption or evolution of heat

– A change in energy occurs as bonds are broken (reactants) and new ones form (products)

– Nearly all chemical reactions absorb or produce heat

– Measured by the heat of reaction or enthalpy

• Enthalpy change is the amount of heat produced or consumed in a process (∆H )

Sign of ∆Hrxn

• Endothermic reactions absorb heat as they occur

– If (∆H ) is positive, then heat is added to the reaction

– If heat supply is removed, the reaction stops

Sign of ∆Hrxn

• Exothermic reactions produce heat as they occur

– If (∆H ) is negative, then heat is evolved by the reaction

Sign of ∆Hrxn

Enthalpy of Reaction

• Photosynthesis reaction– Carbon dioxide reacts with water to produce glucose

and oxygen

• Cell metabolism– Glucose reacts with oxygen to produce carbon dioxide

and water

(g)O6(s)OHC )(OH6(g)CO6 2612622

∆H = +2801 H = +2801 kJkJ

)(OH6(g)CO6 (g)O6(s)OHC 2226126

∆H = -2801 H = -2801 kJkJ

Stoichiometry of ∆Hrxn

• The coefficients in a given chemical reaction represent the number of moles of reactants and products that produce the given heat of reaction (enthalpy change)

• The combustion of methane gas:

• This information gives a quantitative relationship between the heat evolved per mole of methane and oxygen gas

(g)OH 2(g)CO (g)O2(g)CH 2224 ∆HHrxnrxn = -890 = -890 kJkJ

1 mol CH4 = -890 kJ 2 mol O2 = -890 kJ

Stoichiometry of ∆Hrxn

• If the combustion of 1 mol of CH4 with 2 mol O2 releases 890 kJ of heat, the combustion of 2 mol of CH4 with 4 mol of O2 produces twice as much heat

• The equivalence statements can be used to make conversion factors between the amounts of reactants or products and the amount of heat absorbed or emitted in a given reaction

• Calculate the amount of heat emitted when a certain amount in grams undergoes combustion

∆HHrxnrxn = -1780 = -1780 kJkJ

(g)OH 4(g)CO 2 (g)O4(g)CH2 2224

Stoichiometry Involving ∆HH

• Calculate the heat associated with the complete combustion of 4.50 g of methane gas

(g)OH 2(g)CO (g)O2(g)CH 2224 ∆H H rxnrxn= -890 = -890 kJkJ

Given: 4.50 g CH4 Find: kJ

Conversion Factors: 1 mol CH4= -890 kJ 1 mol CH4 = 16.00 g

Solution Map: g CH4 mol CH4 kJ CH4

4

4

00.16

1

CHg

CHmol

41

890

CHmol

kJ

Solution: 450.4 CHg -2.50 × 10-2.50 × 1022 kJ kJ4

4

00.16

1

CHg

CHmol

41

890

CHmol

kJ


• The combustion of sulfur dioxide • It reacts with oxygen to produce sulfur

trioxide

• Calculate the heat produced when 75.2 g of sulfur trioxide is produced

(g)SO2 (g)O(g)SO2 322 ∆H H rxnrxn= -99.1 = -99.1 kJkJ

Given 75.2 g SOGiven 75.2 g SO33

Heat in kJ produced Heat in kJ produced when SOwhen SO3 3 is formedis formed

Find:


Solution Map: Relation between g of SOSolution Map: Relation between g of SO33 and heat and heat released released Grams Grams of SOof SO33

MolesMoles of SOof SO33

kjkj

Write the necessary conversion factors:Write the necessary conversion factors:

3

3

SOmol2

kJ99.1

kJ99.1

SOmol2and

3

3

3

3

SOmol1

SO g80.07

SOg 80.07

SOmol1and

Set up the problem:Set up the problem:

33

33 SOmol2

kJ99.1

SOg80.07

SOmol1SOg75.2 46.5 kJ46.5 kJ

Heat Heat of rxnof rxn

MolarMolarmassmass

33 SOg80.07SOmol1 kJ99.1SOmol2 3

• end

Chapter 8 Quantities In Reactions. Homework Assigned Problems (odd numbers only) “Problems” 17...

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Transcript of Chapter 8 Quantities In Reactions. Homework Assigned Problems (odd numbers only) “Problems” 17...