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Chapter 12

Stoichiometry

12.1 The Arithmetic of Equations

12.2 Chemical Calculations

12.3 Limiting Reagent and

Percent Yield

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What determines how much product you can make?

CHEMISTRY & YOU

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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Limiting and Excess Reagents

How is the amount of product in a reaction affected by an insufficient quantity of any of the reactants?

Limiting and Excess Reagents

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Limiting and Excess Reagents

To make tacos, you need enough meat, cheese, lettuce, tomatoes, sour cream, salsa, and seasonings.

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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.

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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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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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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

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Limiting and Excess Reagents

A balanced chemical equation is a chemist’s recipe.

What would happen if two molecules (moles) of N₂ reacted with three molecules (moles) of H₂?

Experimental Conditions
	Reactants	Products
Before reaction	2 molecules N₂ 3 molecules H₂	0 molecules NH₃

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

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Limiting and Excess Reagents

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.

Experimental Conditions
	Reactants	Products
Before reaction	2 molecules N₂ 3 molecules H₂	0 molecules NH₃
After reaction	1 molecule N₂ 0 molecules H₂	2 molecules NH₃

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Limiting and Excess Reagents

Experimental Conditions
	Reactants	Products
Before reaction	2 molecules N₂ 3 molecules H₂	0 molecules NH₃
After reaction	1 molecule N₂ 0 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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Limiting and Excess Reagents

Experimental Conditions
	Reactants	Products
Before reaction	2 molecules N₂ 3 molecules H₂	0 molecules NH₃
After reaction	1 molecule N₂ 0 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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Limiting and Excess Reagents

Experimental Conditions
	Reactants	Products
Before reaction	2 molecules N₂ 3 molecules H₂	0 molecules NH₃
After reaction	1 molecule N₂ 0 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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What determines how much product you can make in a chemical reaction?

CHEMISTRY & YOU

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What determines how much product you can make in a chemical reaction?

CHEMISTRY & YOU

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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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?

Sample Problem 12.8

Determining the Limiting Reagent in a Reaction

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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

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

UNKNOWN

limiting reagent = ?

Sample Problem 12.8

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Start with one of the reactants and convert from mass to moles.

Calculate Solve for the unknown.

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Sample Problem 12.8

80.0 g Cu × = 1.26 mol Cu

63.5 g Cu

1 mol Cu

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Then, convert the mass of the other reactant to moles.

Calculate Solve for the unknown.

2

Sample Problem 12.8

25.0 g S × = 0.779 mol S

32.1 g S

1 mol S

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Now, convert moles of Cu to moles of S needed to react with 1.25 moles of Cu.

Calculate Solve for the unknown.

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Sample Problem 12.8

1.26 mol Cu × = 0.630 mol S

2 mol Cu

1 mol S

Given quantity

Mole ratio

Needed amount

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Calculate Solve for the unknown.

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Sample Problem 12.8

Compare the amount of sulfur needed with the given amount of sulfur.

0.630 mol S (amount needed to react) <0.779 mol S (given amount)

Sulfur is in excess, so copper is the limiting reagent.

It doesn’t matter which reactant you use. If you used the actual amount of moles of S to find the amount of copper needed, then you would still identify copper as the limiting reagent.

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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.

Evaluate Do the results make sense?

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Sample Problem 12.8

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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)

Sample Problem 12.9

Using Limiting Reagent to Find the Quantity of a Product

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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 = 159.1 g Cu₂S (molar mass)

1 mol Cu₂S/2 mol Cu (mole ratio from balanced equation)

UNKNOWN

Yield = ? g Cu₂S

Sample Problem 12.9

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Start with the moles of the limiting reagent and convert to moles of the product. Use the mole ratio from the balanced equation.

Calculate Solve for the unknown.

2

Sample Problem 12.9

1.26 mol Cu ×

2 mol Cu

1 mol Cu₂S

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Finish the calculation by converting from moles to mass of product.

Calculate Solve for the unknown.

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Sample Problem 12.9

159.1 g Cu₂S

1 mol Cu₂S

1.26 mol Cu ×

2 mol Cu

1 mol Cu₂S

×

= 1.00 × 10² g Cu₂S

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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.

Evaluate Do the results make sense?

3

Sample Problem 12.9

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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?

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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?

Only 2.26 g H₂O are needed to react with 7.0 g Fe. Therefore, Fe 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

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Percent Yield

What does the percent yield of a reaction measure?

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Percent Yield

What does the percent yield of a reaction measure?

A batting average is actually a percent yield.

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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.

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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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Percent Yield

The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent.

Because the actual yield of a chemical reaction is often less than the theoretical yield, the percent yield is often less than 100%.

percent yield =

actual yield

theoretical yield

× 100%

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Percent Yield

The percent yield is a measure of the efficiency of a reaction carried out in the laboratory.

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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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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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Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is

Sample Problem 12.10

Calculating the Theoretical Yield of a Reaction

CaCO₃(s) → CaO(s) + CO₂(g)

Δ

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

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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 = 100.1 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

Sample Problem 12.10

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Start with the mass of the reactant and convert to moles of the reactant.

Calculate Solve for the unknown.

2

Sample Problem 12.10

24.8 g CaCO₃ ×

100.1 g CaCO₃

1 mol CaCO₃

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Next, convert to moles of the product using the mole ratio.

Calculate Solve for the unknown.

2

Sample Problem 12.10

24.8 g CaCO₃ × ×

100.1 g CaCO₃

1 mol CaCO₃

1 mol CaO

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Finish by converting from moles to mass of the product.

Calculate Solve for the unknown.

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Sample Problem 12.10

24.8 g CaCO₃ × × ×

100.1 g CaCO₃

1 mol CaCO₃

1 mol CaO

56.1 g CaO

= 13.9 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.

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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.

Evaluate Does the result make sense?

3

Sample Problem 12.10

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What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO₃ is heated?

Sample Problem 12.11

Calculating the Percent Yield of a Reaction

CaCO₃(s) → CaO(s) + CO₂(g)

Δ

Calculate the theoretical yield first. Then you can calculate the percent yield.

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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 12.10.

UNKNOWN

percent yield = ? %

Sample Problem 12.11

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%

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Substitute the values for actual yield and theoretical yield into the equation for percent yield.

Calculate Solve for the unknown.

2

Sample Problem 12.11

percent yield = × 100% = 94.2%

13.1 g CaO

13.9 g CaO

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In this example, the actual yield is slightly less than the theoretical yield.
Therefore, the percent yield should be slightly less than 100%.

Evaluate Does the result make sense?

3

Sample Problem 12.11

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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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CS₂ + 3Cl₂ → CCl₄ + S₂Cl₂

What is the percent yield of CCl₄ if 617 kg is produced from the reaction of 312 kg of CS₂?

3.12 × 10⁵ g CS₂ × × ×

76.142 g CS₂

1 mol CS₂

1 mol CCl₄

1 mol CS₂

153.81 g CCl₄

1 mol CCl₄

= 6.30 × 10⁵ g CCl₄ = 630 kg CCl₄

Percent yield = × 100% = 97.9%

617 kg CCl₄

630 kg CCl₄

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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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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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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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The Mole and Quantifying Matter

BIG IDEA

The percent yield of a reaction can be calculated from the actual yield and theoretical yield of the reaction.

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END OF 12.3

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