1. Stoichiometry 12.3 Limiting Reagent and Percent Yield Chapter 12
12.1 The Arithmetic of Equations
12.2 Chemical Calculations
12.3 Limiting Reagent and
Percent Yield
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2. What determines how much product you can make?
CHEMISTRY & YOU
What determines how much product you can make?
If a carpenter had two tabletops and seven table legs, he would have difficulty building more than one functional four-legged table.
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3. Limiting and Excess Reagents
To make tacos, you need enough meat, cheese, lettuce, tomatoes, sour cream, salsa, and seasonings.
If you have only 2 taco shells, the quantity of taco shells will limit the number of tacos you can make.
Thus, the taco shells are the limiting ingredient.
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4. Limiting and Excess Reagents
In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.
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5. Limiting and Excess Reagents
A balanced chemical equation is a chemist’s recipe.
Chemical Equations
N2(g) +
3H2(g)
2NH3(g)
“Microscopic recipe”
1 molecule N2
3 molecules H2
2 molecules NH3
“Macroscopic recipe”
1 mol N2
3 mol H2
2 mol NH3
What would happen if two molecules (moles) of N2 reacted with three molecules (moles) of H2?
Experimental Conditions
Reactants
Products
Before reaction
2 molecules N molecules H2
0 molecules NH3
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6. Limiting and Excess Reagents
Experimental Conditions
Reactants
Products
Before reaction
2 molecules N molecules H2
0 molecules NH3
After reaction
1 molecule N molecules H2
2 molecules NH3
Before the reaction takes place, N2 and H2 are present in a 2:3 molecule (mole) ratio.
As the reaction takes place, one molecule (mole) of N2 reacts with 3 molecules (moles) of H2 to produce two molecules (moles) of NH3.
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7. Limiting and Excess Reagents
Experimental Conditions
Reactants
Products
Before reaction
2 molecules N molecules H2
0 molecules NH3
After reaction
1 molecule N molecules H2
2 molecules NH3
All the H2 has now been used up, and the reaction stops.
One molecule (mole) of unreacted N2 is left in addition to the two molecules (moles) of NH3 that have been produced by the reaction.
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8. Limiting and Excess Reagents
Experimental Conditions
Reactants
Products
Before reaction
2 molecules N molecules H2
0 molecules NH3
After reaction
1 molecule N molecules H2
2 molecules NH3
In this reaction, only the hydrogen is completely used up.
H2 is the limiting reagent, or the reactant that determines the amount of product that can be formed by a reaction.
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9. Limiting and Excess Reagents
Experimental Conditions
Reactants
Products
Before reaction
2 molecules N molecules H2
0 molecules NH3
After reaction
1 molecule N molecules H2
2 molecules NH3
The reactant that is not completely used up in a reaction is called the excess reagent.
In this example, nitrogen is the excess reagent because some nitrogen remains unreacted.
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10. What determines how much product you can make in a chemical reaction?
CHEMISTRY & YOU
What determines how much product you can make in a chemical reaction?
A limited quantity of any of the reactants that are needed to make a product will limit the amount of product that forms.
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11. Sample Problem 12.8
Determining the Limiting Reagent in a Reaction
Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation:
2Cu(s) + S(s)  Cu2S(s)
What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?
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12. Analyze List the knowns and the unknown.
Sample Problem 12.8
Analyze List the knowns and the unknown. 1
The number of moles of each reactant must first be found. The balanced equation is used to calculate the number of moles of one reactant needed to react with the given amount of the other reactant.
KNOWNS
UNKNOWN
mass of copper = 80.0 g Cu
mass of sulfur = 25.0 g S
molar mass of Cu = 63.5 g/mol
molar mass of S = 32.1 g/mol
1 mol S/2 mol Cu
limiting reagent = ?
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13. Sample Problem 12.9
Using Limiting Reagent to Find the Quantity of a Product
What is the maximum number of grams of Cu2S that can be formed when 80.0 g Cu reacts with 25.0 g S?
2Cu(s) + S(s)  Cu2S(s)
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14. Analyze List the knowns and the unknown.
Sample Problem 12.9
Analyze List the knowns and the unknown. 1
The limiting reagent, which was determined in the previous sample problem, is used to calculate the maximum amount of Cu2S formed.
KNOWNS
limiting reagent = 1.26 mol Cu (from sample problem 12.8)
1 mol Cu2S = g Cu2S (molar mass)
1 mol Cu2S/2 mol Cu (mole ratio from balanced equation)
UNKNOWN
Yield = ? g Cu2S
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15. Calculate Solve for the unknown.
Sample Problem 12.9
Calculate Solve for the unknown. 2
Finish the calculation by converting from moles to mass of product.
159.1 g Cu2S
1 mol Cu2S
1.26 mol Cu 
2 mol Cu 
= 1.00  102 g Cu2S
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16. Evaluate Do the results make sense?
Sample Problem 12.9
Evaluate Do the results make sense? 3
Copper is the limiting reagent in this reaction. The maximum number of grams of Cu2S produced should be more than the amount of copper that initially reacted because copper is combining with sulfur.
However, the mass of Cu2S produced should be less than the total mass of the reactants (105.0 g) because sulfur was in excess.
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17. Rust forms when iron, oxygen, and water react
Rust forms when iron, oxygen, and water react. One chemical equation for the formation of rust is
2Fe + O2 + 2H2O  2Fe(OH)2
If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent?
7.00 g Fe    = 2.26 g H20
1 mol Fe
55.85 g Fe
18.0 g H2O
1 mol H2O
2 mol Fe
2 mol H2O
Only 2.26 g H2O are needed to react with 7.0 g Fe. Therefore, Fe is the limiting reagent.
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18. Percent Yield Percent Yield
What does the percent yield of a reaction measure?
A batting average is actually a percent yield.
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19. Percent Yield
When a balanced chemical equation is used to calculate the amount of product that will form during a reaction, the calculated value represents the theoretical yield.
The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants.
The amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield.
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20. Percent Yield
The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent.
percent yield =
actual yield
theoretical yield
 100%
Because the actual yield of a chemical reaction is often less than the theoretical yield, the percent yield is often less than 100%.
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21. Percent Yield
The percent yield is a measure of the efficiency of a reaction carried out in the laboratory.
The mass of the reactant is measured.
The mass of one of the products, the actual yield, is measured. The percent yield is calculated.
The reactant is heated.
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22. Percent Yield Many factors cause percent yields to be less than 100%.
Reactions do not always go to completion; when a reaction is incomplete, less than the calculated amount of product is formed.
Impure reactants and competing side reactions may cause unwanted products to form.
Actual yield can be lower than the theoretical yield due to a loss of product during filtration or in transferring between containers.
If reactants or products have not been carefully measured, a percent yield of 100% is unlikely.
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23. CaCO3(s)  CaO(s) + CO2(g)
Sample Problem 12.10
Calculating the Theoretical Yield of a Reaction
Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is
CaCO3(s)  CaO(s) + CO2(g) D
What is the theoretical yield of CaO if 24.8 g CaCO3 is heated?
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24. Analyze List the knowns and the unknown.
Sample Problem 12.10
Analyze List the knowns and the unknown. 1
Calculate the theoretical yield using the mass of the reactant.
KNOWNS
mass of CaCO3 = 24.8 g CaCO3
1 mol CaCO3 = g CaCO3 (molar mass)
1 mol CaO = 56.1 g CaO (molar mass)
1 mol CaO/1 mol CaCO3 (mole ratio from balanced equation)
UNKNOWN
theoretical yield = ? g CaO
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25. Calculate Solve for the unknown.
Sample Problem 12.10
Calculate Solve for the unknown. 2
Finish by converting from moles to mass of the product.
24.8 g CaCO3   
100.1 g CaCO3
1 mol CaCO3
1 mol CaO
56.1 g CaO
If there is an excess of a reactant, then there is more than enough of that reactant and it will not limit the yield of the reaction.
= 13.9 g CaO
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26. Evaluate Does the result make sense?
Sample Problem 12.10
Evaluate Does the result make sense? 3
The mole ratio of CaO to CaCO3 is 1:1. The ratio of their masses in the reaction should be the same as the ratio of their molar masses, which is slightly greater than 1:2.
The result of the calculations shows that the mass of CaO is slightly greater than half the mass of CaCO3.
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27. CaCO3(s)  CaO(s) + CO2(g)
Sample Problem 12.11
Calculating the Percent Yield of a Reaction
What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO3 is heated?
Calculate the theoretical yield first. Then you can calculate the percent yield.
CaCO3(s)  CaO(s) + CO2(g) D
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28. Analyze List the knowns and the unknown.
Sample Problem 12.11
Analyze List the knowns and the unknown. 1
Use the equation for percent yield. The theoretical yield for this problem was calculated in Sample Problem
KNOWNS
actual yield = 13.1 g CaO
theoretical yield = 13.9 g CaO (from sample problem 12.10)
percent yield =
actual yield
theoretical yield
 100%
UNKNOWN
percent yield = ? %
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29. Calculate Solve for the unknown.
Sample Problem 12.11
Calculate Solve for the unknown. 2
Substitute the values for actual yield and theoretical yield into the equation for percent yield.
percent yield =  100% = 94.2%
13.1 g CaO
13.9 g CaO
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30. Evaluate Does the result make sense?
Sample Problem 12.11
Evaluate Does the result make sense? 3
In this example, the actual yield is slightly less than the theoretical yield.
Therefore, the percent yield should be slightly less than 100%.
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31. Carbon tetrachloride, CCl4, is a solvent that was once used in large amounts in dry cleaning. One reaction that produces carbon tetrachloride is
CS2 + 3Cl2  CCl4 + S2Cl2
What is the percent yield of CCl4 if 617 kg is produced from the reaction of 312 kg of CS2?
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32. What is the percent yield of CCl4 if 617 kg is produced from the reaction of 312 kg of CS2?
CS2 + 3Cl2  CCl4 + S2Cl2
3.12  105 g CS2   
g CS2
1 mol CS2
1 mol CCl4
g CCl4
= 6.30  105 g CCl4 = 630 kg CCl4
Percent yield =  100% = 97.9%
617 kg CCl4
630 kg CCl4
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33. Key Concepts and Key Equation
In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.
The percent yield is a measure of the efficiency of a reaction performed in the laboratory.
percent yield =
actual yield
theoretical yield
 100%
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34. Glossary Terms
limiting reagent: any reactant that is used up first in a chemical reaction; it determines the amount of product that can be formed in the reaction
excess reagent: a reagent present in a quantity that is more than sufficient to react with a limiting reagent; any reactant that remains after the limiting reagent is used up in a chemical reaction
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35. Glossary Terms
theoretical yield: the amount of product that could form during a reaction calculated from a balanced chemical equation; it represents the maximum amount of product that could be formed from a given amount of reactant
actual yield: the amount of product that forms when a reaction is carried out in the laboratory
percent yield: the ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percentage; a measure of the efficiency of a reaction
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36. The Mole and Quantifying Matter
BIG IDEA
The Mole and Quantifying Matter
The percent yield of a reaction can be calculated from the actual yield and theoretical yield of the reaction.
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