Part 7 Stoichiometry

8/12/2019 Part 7 Stoichiometry

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Atomic Masses; the Carbon-12 Scale

Individual atoms are far too small to be weighed on a balance.

Atomic massis the mass of an atom in atomic mass units

(amu).

The atomic mass of an element indicates how heavy, on the

average, one atom of that element is compared with an atom

of another element.

To set up a scale of atomic mass, it is needed a standard valuefor one particular species.

One atomic mass unit is defined as a mass exactly equal to

one-twelfth the mass of one carbon-12 atom.


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On this scale1H = 1.008 amu

16O = 16.00 amu

1 amu = 1/12 x mass of one C-12 atom

In the periodic table, atomic masses are

listed directly below the symbol of the

element


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Relative Atomic Mass, Ar

; Relative Molecule Mass, Mr

Some definitions of atomic and molecular masses:

If the mass ratio of one atom of an element to one atom of

carbon-12 is known, the relative atomic mass of the element can

be calculated.

Relative atomic mass, Arof an =

element X

Mass of one atom of X

1/12 x mass of one 12C atom

Relative molecular mass, Mrof a =

compound W

Mass of one molecule of W



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The mass ratio of one atom of nitrogen to one atom of

carbon-12 is 1.24899. What is the relative mass of

nitrogen?

Start from the definition of relative atomic mass of an element

= 12 x 1.24899

= 1.24899Mass of one atom of N

mass of one atom of carbon-12

Relative atomic mass of N = Mass of one atom of N


= 14.98788


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The amount of each isotope of an element expressed inpercentage is called the relative abundance or isotopic

abundance of that element.

The total relative abundance is 100%.

Element that have more than one isotopemeasure theaverage mass for all occurring mixture of isotopes.

If an element has n isotopes, the average atomic mass of the

elements is:

Fraction abundance,f =Percentage of isotope

100


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Example: Lithium

(7.42 x 6.015) + (92.58 x 7.016)

100= 6.941 amu


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Tentukan jisim atom relatif, Arbagi unsur Neon yangdiketahui mempunyai tiga isotop iaitu 20Ne, 21Ne dan 22Ne

dengan peratus kelimpahan masing-masing 90. 92%, 0.26%

dan 8.82%.

= 20.18

90.92

100+Ar Ne = ( x 20 ) ( )x 21

0.26

100

( )x 228.82

100

+


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The mole(mol) is the amount of a substance that contains

as many elementary entities as there are atoms in exactly

12.00 grams of 12C.

The definition specifies the number of objects in a fixedmass of substance.

Therefore, 1 mole of substance = fixed number of chemical

entities and has a fixed mass.


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Water

18.02 gCaSO4.2H2O172.19 g

Oxygen

32.00 g

Copper

63.55 g

One mole of

common

substances.


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Molar massis the mass of 1 mole of in grams

ions

Atoms

Molecules

1 mole 12C atoms = 6.022 x 1023atoms = 12.00 g

1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li


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Mass (g) = no. of moles x no of grams

1 mol

No. of moles = mass (g) x 1 mol

no. of grams

Convert between amount (mol) and mass (g), use molarmass ( M in g/mol)

No. of entities = no. of moles x 6.022 x 1023entities

no. of grams

No. of moles = no. of entities x 1 mol

6.022 x 1023entities

Convert between amount (mol) and no. of entities, use

Avogadros number (6.022 x 1023

entities)


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Convert between number of entities and mass, first

convert to number of moles.

No. of atoms = mass (g) x 1 mol x 6.022 x 1023entities

no. of grams 1 mol


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Summary of the

mass-mole-number

relationships for

elements.


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Acetylsalicylic acid, C9H8O4, is the active ingredient of aspirin.

a.What is the mass in grams of 0.509 mol of Acetylsalicylic

acid?b.How many moles of C9H8O4are in 1 g sample of aspirin that

contains 91.6% by mass of C9H8O4?

= [9(12.01) + 8(1.008) + 4(16.00)] g/mol = 180.15 g/mol

The molar mass of C9H8O4

a. 0.509 mol x 180.15 g = 91.7g

1 molb. 1.000 g x 0.916 x 1 mol = 5.08 x 10-3mol

180.15 g


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Do You Understand Molar Mass?

How many atoms are in 0.551 g of potassium (K) ?

1 mol K = 39.10 g K

1 mol K = 6.022 x 1023atoms K

0.551 g K1 mol K

39.10 g Kx x

6.022 x 1023atoms K

1 mol K

= 8.49 x 1021atoms K


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Sample Problem Calculating Number of Atoms in a Given Mass of

an Element

PROBLEM: Iron (Fe), the main component of steel, is the most important metal

in industrial society. How many Fe atoms are in 95.8 g of Fe?

To convert g of Fe to atoms, first find the # mols of Fe and then convert

mols to atoms.

PLAN:

SOLUTION:

95.8 g Fe x55.85 g Fe

mol Fe= 1.72 mol Fe

1.72mol Fe x

6.022x1023 atoms Fe

mol Fe


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Molecular mass(or molecular weight) is the sum of theatomic masses (in amu) in a molecule.

SO2

1S 32.07 amu

2O + 2 x 16.00 amuSO2 64.07 amu

1 molecule SO2= 64.07 amu

1 mole SO2= 64.07 g SO2


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Do You Understand Molecular Mass?

How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol H = 6.022 x 1023atoms H

= 5.82 x 1024atoms H

1 mol C3

H8

O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms

1 mol C3H8Ox

6.022 x 1023H atoms

1 mol H atomsx

Gram of C3H8O mol of C3H8O mol of H Atom of H


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Sample ProblemCalculating the Moles and Number of Formula Units in a

Given Mass of a Compound

PROBLEM:

PLAN:

SOLUTION:

How many formula units are in 41.6 g of ammonium

carbonate?Write the formula for the compound. Calculate M. Convert the givenmass to mols and then mols to formula units.

The formula is (NH4)2CO3

M = (2 x 14.01 g/mol N) + (8 x 1.008 g/mol H)

+ (12.01 g/mol C) + (3 x 16.00 g/mol O) = 96.09 g/mol

41.6 g (NH4)2CO3 xmol (NH4)2CO3

96.09 g (NH4)2CO3

6.022x1023formula units (NH4)2CO3

mol (NH4)2CO3x


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nX molar mass of element

molar mass of compoundx 100%

nis the number of moles of the element in 1 mole of the compound

C2

H6

O

%C =2x (12.01 g)

46.07 gx 100% = 52.14%

%H = 6x (1.008 g)46.07 g

x 100% = 13.13%

%O =1x (16.00 g)

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

Percent composition by mass is the percent by mass of each

element in a compound


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Percent Composition and Empirical Formulas

Determine the empirical formula of a

compound that has the following percentcomposition by mass:

K 24.75, Mn 34.77, O 40.51 percent.

nK= 24.75 g K x = 0.6330 mol K1 mol K

39.10 g K

nMn= 34.77 g Mn x = 0.6329 mol Mn1 mol Mn

54.94 g Mn

nO= 40.51 g O x = 2.532 mol O1 mol O

16.00 g O


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K : ~~ 1.00.6330

0.6329

Mn :0.6329

0.6329= 1.0

O :~~ 4.0

2.532

0.6329

nK= 0.6330, nMn= 0.6329, nO= 2.532

Percent Composition and Empirical Formulas


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Empirical formulathe simplest whole-number ratioof moles of each element in the compound.

Example: Ethanol


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g of O = g of sample(g of C + g of H) = 4.0 g O

g CO2 mol CO2 mol C g C = 6.0 g C

22.0 g CO21 mol CO2

44.01 g CO2x

1 mol C atoms

1 mol CO2x

12.01 C atoms

1 mol C atomsx

g H2O mol H2O mol H g H = 1.5 g H

13.5 g H2O1 mol H2O

18.02 g H2O

x1 mol H atoms

1 mol H2O

x1.008 H atoms

1 mol H atoms

x


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The number of moles of each element in 11.5 g of

ethanol:

Empirical formula C0.5H1.5O0.25

Divide by smallest subscript (0.25)

Moles of C = 6.00 g C x 1 mol C = 0.50 mol C

12.01 g C

Moles of H = 1.51 g H x 1 mol C = 1.50 mol H1.008 g C

Moles of O = 4.00 g O x 1 mol C = 0.25 mol O

16.00 g C


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Example: Ascorbic acid composed 40.92% carbon, 4.58%hydrogen and 54.50% oxygen by mass. Determine its

empirical formula.

If we have 100 g of ascorbic acid, then % can be converted

to grams.

Moles of C = 40.92 g C x 1 mol C = 3.407 mol C

12.01 g C

Moles of H = 4.58 g H x 1 mol C = 4.54 mol H1.008 g C

Moles of O = 54.5 g O x 1 mol C = 3.406 mol O

16.00 g C

Empirical formula C3.407H4.54O3.406


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Example: Ascorbic acid composed 40.92% carbon, 4.58%hydrogen and 54.50% oxygen by mass. Determine its

empirical formula.

Convert the whole number by dividing to smaller number

C : 3.407 1

3.406

H : 4.54 1.33

3.406

O : 3.406 1

3.406

1.33 x 1 = 1.33

1.33 x 2 = 2.66

1.33 x 3 = 3.99 4


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Molecular formula can be obtain by using empiricalformulathe actual number of moles of each

element in 1 mol of compound.

Example: Hydrogen peroxide Empirical formula HO (17.01 g/mol)

Molecular formula H2O2 (34.02 g/mol)

Whole no. multiple = molar mass (g/mol)empirical formula mass (g/mol)

= 34.02g/mol = 2.000 = 2

17.01 g/mol


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Example: 3 ways of representing the reaction of H2with O2to form

H2O


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2 atoms Mg + 1 molecule O2makes 2 formula units MgO

2 moles Mg + 1 mole O2makes 2 moles MgO

48.6 grams Mg + 32.0 grams O2makes 80.6 g MgO

IS NOT

2 grams Mg + 1 gram O2makes 2 g MgO


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1. Write the correctformula(s) for the reactants on the leftside and the correctformula(s) for the product(s) on the

right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and water

C2H6 + O2 CO2 + H2O

2. Change the numbers in front of the formulas (coefficients)

to make the number of atoms of each element the sameon both sides of the equation. Do not change the

subscripts.

2C2H6 NOT C4H12


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C2H6+ O2 CO2+ H2O start with C or H but not O

2 carbonon left

1 carbonon right

multiply CO2by

2

C2H6+ O2 2CO2+ H2O

6 hydrogen

on left

2 hydrogen

on rightmultiply H2O by 3

3. Start by balancing those elements that appear in only one

reactant and one product.


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4. Balance those elements that appear in two or morereactants or products.

multiply O2by72

2 oxygen

on left

4 oxygen

(2x2)

C2H6+ O2 2CO2+ 3H2O

+ 3 oxygen

(3x1)

= 7 oxygen

on right

C2H6+ O2 2CO2+ 3H2O72

remove fraction

multiply both sides by 2

2C2H6+ 7O2 4CO2+ 6H2O


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5. Check to make sure that you have the same number ofeach type of atom on both sides of the equation.

2C2H6+ 7O2 4CO2+ 6H2O

Reactants Products

4 C12 H

14 O

4 C12 H

14 O

4 C (2x 2) 4 C12 H (2x 6) 12 H (6x 2)14 O (7x 2) 14 O (4x 2 + 6)


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translate the statement

Sample Problem Balancing Chemical Equations

PROBLEM:

PLAN: SOLUTION:

balance the atoms

Within the cylinders of a cars engine, the hydrocarbon octane

(C8H18), one of many components of gasoline, mixes with oxygen

from the air and burns to form carbon dioxide and water vapor.

Write a balanced equation for this reaction.

adjust the coefficients

check the atom balance

C8H18+ O2 CO2+ H2O

C8H18+ O2 CO2+ H2O825/2 9

2 C8H18+ 25 O2 16 CO2+ 18H2O

2C8H18+ 25O2 16 CO2+ 18 H2O

2 C8H18(l) + 25 O2 (g) 16 CO2 (g) + 18 H2O (g)


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Stoichiometrically Equivalent Molar ratios

Stoichiometry is the quantitative study of reactants and

products in a chemical reaction.

Use moles to calculate the amount of product formed in

a reactionmole method.

Therefore, the stoichiometric coefficients in a chemical

equation can be interpreted as the number of moles of

each substance.


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Example:

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

The equation can be read as 1 mole of N2gas combines

with 3 moles of H2gas to form 2 moles of NH3gas.

In stoichiometric calculation 3 moles of H2are stoichiometrically equivalent to 2 moles of NH3

1 mole N2are stoichiometrically equivalent to 2 mol NH3

1 mole N2are stoichiometrically equivalent to 3 mol H2

The conversion factors from this equivalence:

1 molecule 3 molecule 2 molecule

3 mol H2

2 mol NH3

2 mol NH3

3 mol H2

and


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Example: 6.0 moles of H2react completely with N2to form

NH3. Calculate the amount of NH3produces in moles

Moles of NH3produced = 6.0 mol H2 x 2 mol NH3

3 mol H2

= 4.0 mol NH3 Example: 16.0 g of H2react completely with N2to form

NH3. How many grams of NH3will be formed?

Conversion

factor

grams H2 moles of H2 moles of NH3 grams of NH3

16 g H21 mol H22.016 g H2

x 2 mol NH33 mol H2

x 17.03 g NH31 mol NH3

x

Grams of NH3=

= 90.1 g NH3


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The general approach for solving stoichiometryproblems:

i. Write a balanced equation for the reaction.

ii. Convert the given amount of the reactant (gram @other unit) to number of moles.

iii. Use the mole ratio from the balanced equation to

calculate the number of moles of product formed.

iv. Convert the moles of product to gram ( or otherunits) of product.


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Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2+ 4H2O

If 209 g of methanol are used up in the combustion,

what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2Omolar mass

CH3OH

coefficients

chemical equation

molar mass

H2O

209 g CH3OH 1 mol CH3OH32.0 g CH3OHx 4 mol H2O2 mol CH3OH

x 18.0 g H2O1 mol H2Ox

= 235 g H2O


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Sample Problem Calculating Amounts of Reactants and Products

PROBLEM: Copper is obtained from copper(I) sulfide by roasting it in the

presence of oxygen gas) to form powdered copper(I) oxide and

gaseous sulfur dioxide.

PLAN:

find mols O2 find mols SO2

find g SO2

find mols Cu2O

find mols O2

find kg O2

a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?

b)How many grams of sulfur dioxide are formed when 10.0 mol of copper(I)

sulfide is roasted?

c)How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?



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SOLUTION:


2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

= 0.959 kg O2kg O2

103 g O2

20.0 mol Cu2O3 mol O2

2 mol Cu2O

32.00 g O2

mol O2

3 mol O22 mol Cu2S

= 15.0 mol O210.0 mol Cu2S

= 641 g SO210.0 mol Cu2S

2 mol SO2

2 mol Cu2S

64.07 g SO2

mol SO2

= 20.0 mol Cu2O2.86 kg Cu2O 10

3 g Cu2O

kg Cu2O

mol Cu2O

143.10 g Cu2O

(a)How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?

(b)How many grams of sulfur dioxide are formed when 10.0 mol of copper(I)

sulfide is roasted?

c)How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?


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Goal of reaction is to produce the maximum quantityfrom the starting material.

Consequently, some reactant will be left over at the end

of reaction.

The reactant used up first in a reaction is called thelimiting reagent- the maximum amount of product

formed depends on how much of this reactant was

originally present.

Excess reagents are the reactants present in quantities

greater than necessary to react with the quantity of the

limiting reagent.


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6 green used up6 red left over


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Example: Urea is prepares by reacting ammonia with CO22NH3(g) + CO2(g) (NH2)2CO (aq) + H2O(l)

637.2 g of NH3are treated with 1142 g CO2. Which of the

two reactant is limiting reagent?

grams NH3 moles of NH3 moles of (NH2)2CO

637.2 g NH31 mol NH317.03 g NH3

x 1 mol (NH2)2CO2 mol NH3

x

moles of (NH2

)2

CO

=

= 18.71 mol (NH2)2CO


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637.2 g of NH3are treated with 1142 g CO2. Which of the two

reactant is limiting reagent?

grams CO2 moles of CO2 moles of (NH2)2CO

1142 g CO2 1 mol CO244.01 g CO2

x 1 mol (NH2)2CO1 mol CO2

x

moles of (NH2)2CO =

= 25.95 mol (NH2)2COTherefore, NH3must be the limiting reagent because it produces a

smaller amount of (NH2)2CO


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Calculate the mass of (NH2)2CO formed.

18.71 mol (NH2)2CO60.06 g (NH2)2CO1 mol (NH2)2CO

x

Mass of (NH2)2CO

=

= 1124 g (NH2)2CO

The molar mass of (NH2)2CO is 60.06 g. Use the conversion

factorto convert from moles of (NH2)2CO to gram of

(NH2)2CO.


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Do You Understand Limiting Reagents?

In one process, 124 g of Al are reacted with 601 g of Fe2

O3

2Al + Fe2O3 Al2O3+ 2Fe

Calculate the mass of Al2O3formed.

g Al mol Al mol Fe2

O3

needed g Fe2

O3

needed

or

g Fe2O3 mol Fe2O3 mol Al needed g Al needed

124 g Al 1 mol Al27.0 g Al

x 1 mol Fe2O32 mol Al

x 160. g Fe2O31 mol Fe2O3

x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent


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Use limiting reagent (Al) to calculate amount of product thatcan be formed.

g Al mol Al mol Al2O3 g Al2O3

124 g Al1 mol Al

27.0 g Alx

1 mol Al2O3

2 mol Alx

102. g Al2O3

1 mol Al2O3x

= 234 g Al2O3

Do You Understand Limiting Reagents?

In one process, 124 g of Al are reacted with 601 g of Fe2

O3

2Al + Fe2O3 Al2O3+ 2Fe

Calculate the mass of Al2O3formed.


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Theoretical Yieldis the amount of product that wouldresult if all the limiting reagent reacted (from calculation

using the balanced equation).

Actual Yieldis the amount of product actually obtained

from a reaction (almost always less than the theoreticalyield).

Percent yield - to determine reaction efficiency, which

describe the proportion of the actual yield to the

theoretical yield.

% Yield=Actual Yield

Theoretical Yieldx 100


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Example:

TiCl4(g) + 2Mg(l) Ti(s) + 2MgCl2 (l)

3.54 x 107g of TiCl4are reacted with 1.13 x 107g of Mg.

Calculate the theoretical yield of Ti in grams.

g of TiCl4 Mol of TiCl4 Mol of Ti

First: to find limiting reagent, calculate the no. of moles of Ti that

could produces if all the TiCl4reacted

= 1.87 x 105mol Ti

Next: calculate the no. of moles of Ti formed from 1.13 x 107g

of Mg.

g of Mg Mol of Mg Mol of Ti = 2.32 x 105mol Ti


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Example:



Calculate the theoretical yield of Ti in grams.

Therefore, TiCl4is the limiting reagent because it produces a

smaller amount of Ti

= 8.95 x 106g Ti1.87 x 105mol Ti47.88 g Ti

1 mol Tix

Mass of Ti =


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Example:



Calculate the percent yield if 7.91 x 106g of Ti are

actually obtained.

The percent yield is given by:

% Yield =Actual Yield

Theoretical Yield

x 100

% Yield =7.91 x 106g

8.95 x 106gx 100

= 88.4%


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Sample Problem Calculating Percent Yield

PROBLEM: Silicon carbide (SiC) is made by reacting sand(silicon dioxide,

SiO2) with powdered carbon at high temperature. Carbon

monoxide is also formed. What is the percent yield if 51.4 kg

of SiC is recovered from processing 100.0 kg of sand?

PLAN: SOLUTION:

SiO2(s) + 3C(s) SiC(s) + 2CO(g)

103 g SiO2

kg SiO2100.0 kg SiO2

mol SiO2

60.09 g SiO2= 1664 mol SiO2

mol SiO2= mol SiC = 1664

1664 mol SiC 40.10 g SiCmol SiC

kg103g

= 66.73 kg

x 100 % = 77.0%51.4 kg

66.73 kg


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Part 7 Stoichiometry

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