chapter 11 stoichiometry 11.1 defining stoichiometry · pdf filechapter 11 stoichiometry...

CHAPTER 11 Stoichiometry

Chemistry: Matter and Change Solving Problems: A Chemistry Handbook

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11.1 Defining Stoichiometry

Stoichiometry is the study of quantitative relationships between amounts

of reactants used and products formed by a chemical reaction.

Stoichiometry is based on the law of conservation of mass. The law of

conservation of mass states that matter is neither created nor destroyed.

Thus, in a chemical reaction, the mass of the reactants equals the mass

of the products. You can use stoichiometry to answer questions about

the amounts of reactants used or products formed by reactions. For

example, look at the balanced chemical equation for the formation of

table salt (NaCl).

2Na(s) Cl2(g) 2NaCl(s)

You could use stoichiometry to answer the following questions about the

chemical reaction.

How much sodium is needed to produce a certain amount of

table salt?

How much chlorine is needed to produce a certain amount of

table salt?

Given a certain amount of sodium or chlorine, how much table

salt can be produced?

When you look at a balanced equation, there are two ways to interpret

the coefficients. The coefficients tell you how many individual particles

are interacting in the chemical reaction. For example, from the chemical

equation above, you learn that two sodium atoms react with one chlorine

molecule to form two formula units of table salt. You also learn that two

moles of sodium react with one mole of chlorine to form two moles of

salt. What the coefficients do not tell you directly is the masses of the

reactants and the products in the chemical reaction.

Example Problem 1 Interpreting Chemical Equations

Interpret the balanced chemical equation in terms of particles, moles, and

mass. Show that the law of conservation of mass is observed.

Chapter 11 (continued)


2

4NH3 5O2 4NO 6H2O

The coefficients represent both the numbers of particles and the numbers

of moles interacting in the chemical reaction.

4 molecules NH3 5 molecules O2

4 molecules NO 6 molecules H2O

4 moles NH3 + 5 moles O2 4 moles NO + 6 moles H2O

You can calculate the mass of each reactant and product by multiplying

the number of moles by the conversion factor molar mass.

4 mol NH3

17.03 g NH3

1 mol NH3

68.12 g NH3

25 mol O 2

2

32.00 g O

1 mol O 2160.0 g O

4 mol NO 30.01 g NO

1 mol NO 120.0 g NO

6 mol H2O

18.02 g H2O

1 mol H2O

108.1 g H2O

The law of conservation of mass is observed because the mass of the

reactants (68.12 g NH3 + 160.0 g O2 = 228.1 g) equals the mass of the

products (120.0 g NO + 108.1 g H2O 228.1 g).

Practice Problems

1. Interpret each balanced equation in terms of particles, moles, and

mass. Show that the law of conservation of mass is observed.

a. 2H2O2(l) O2(g) 2H2O(l)

b. H2CO3(aq) H2O(l) CO2(g)

c. 4HCl(aq) O2(g) 2H2O(l) 2Cl2(g)

As you know, the coefficients in a balanced chemical equation indicate

the relationships among the moles of reactants and products in the

reaction. You can use the coefficients to write mole ratios. A mole ratio

is a ratio between the numbers of moles of any two substances in a

balanced chemical equation. What mole ratios can be written for the

following chemical equation?



3

2H2O2(l) O2(g) 2H2O(l)

2 mol H2O

2

1 mol O2

and 2 mol H

2O

2

2 mol H2O

1 mol O2

2 mol H2O

2

and 1 mol O

2

2 mol H2O

2 mol H2O

2 mol H2O

2

and 2 mol H

2O

1 mol O2

To determine the number of mole ratios that defines a given chemical

reaction, multiply the number of species in the equation by the next

lower number. Thus, a chemical reaction with three participating species

can be defined by six mole ratios (3 2 6); a chemical reaction with

four species can be defined by 12 mole ratios (4 3 12).

Why learn to write mole ratios? They are the key to calculations that

are based on chemical equations. Using a balanced chemical equation,

mole ratios derived from the equation, and a given amount of one of the

reactants or products, you can calculate the amount of any other

participant in the reaction.

Practice Problems

2. Determine all the mole ratios for the following balanced chemical

equations.

a. N2(g) O2(g) 2NO(g)

b. 4NH3(aq) 5O2(g) 4NO(g) 6H2O(l)


11.2 Stoichiometric Calculations

There are three basic stoichiometric calculations: mole-to-mole

conversions, mole-to-mass conversions, and mass-to-mass conversions.

All stoichiometric calculations begin with a balanced equation and mole

ratios.

Stoichiometric mole-to-mole conversion How can you

determine the number of moles of table salt (NaCl) produced from

0.02 moles of chlorine (Cl2)?



4

First, write the balanced equation.

2Na(s) Cl2(g) 2NaCl(s)

Then, use the mole ratio to convert the known number of moles of

chlorine to the number of moles of table salt. Use the formula below.

moles of known

moles of unknown

moles of known moles of unknown

0.02 mol Cl2

2 mol NaCl

1 mol Cl2

0.04 mol Cl2

Example Problem 2 Stoichiometric Mole-to-Mole Conversion

A piece of magnesium burns in the presence of oxygen, forming

magnesium oxide (MgO). How many moles of oxygen are needed to

produce 12 moles of magnesium oxide?

Write the balanced equation and the mole ratio that relates mol

O2 to mol MgO.

2Mg(s) + O2(g) 2MgO(s)

1 mol O2

2 mol MgO

Multiply the known number of moles of MgO by the mole ratio.

12 mol MgO 1 mol O

2

2 mol MgO 6 mol O

2

Six moles of oxygen is needed to produce 12 moles of magnesium oxide.

Practice Problems

3. The carbon dioxide exhaled by astronauts can be removed from a

spacecraft by reacting it with lithium hydroxide (LiOH). The

reaction is as follows: CO2(g) 2LiOH(s) Li2CO3(s) H2O(l).

An average person exhales about 20 moles of CO2 per day. How

many moles of LiOH would be required to maintain two astronauts

in a spacecraft for three days?



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4. Balance the following equation and answer the questions below.

KClO3(s) KCl(s) O2(g)

a. How many moles of O2 are produced from 10 moles of KClO3?

b. How many moles of KCl are produced using 3 moles of KClO3?

c. How many moles of KClO3 are needed to produce 50 moles

of O2?

Stoichiometric mole-to-mass conversion A mole-to-mass

conversion allows you to calculate the mass of a product or reactant in a

chemical reaction given the number of moles of a reactant or product.

Example Problem 3 Stoichiometric Mole-to-Mass Conversion

The following reaction occurs in plants undergoing photosynthesis.

6CO2(g) 6H2O(l) C6H12O6(s) 6O2(g)

How many grams of glucose (C6H12O6) are produced when 24.0 moles of

carbon dioxide reacts in excess water?

Determine the number of moles of glucose produced by the given

amount of carbon dioxide.

24.0 mol CO2

1 mol C6H

12O

6

6 mol CO2

4.00 mol C6H

12O

6

Multiply by the molar mass.

4.00 mol C6H

12O

6

180.18 g C6H

12O

6

1 mol C6H

12O

6

721 g C6H

12O

6

721 grams of glucose is produced from 24.0 moles of carbon dioxide.

Practice Problems

5. Calculate the mass of sodium chloride (NaCl) produced when

5.50 moles of sodium reacts in excess chlorine gas.



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6. How many grams of chlorine gas must be reacted with excess

sodium iodide (NaI) to produce 6.00 moles of sodium chloride?

a. Balance the equation: NaI(aq) Cl2(g) NaCl(aq) I2(s).

b. Perform the calculation.

7. Calculate the mass of hydrochloric acid (HCl) needed to react with

5.00 moles of zinc.

a. Balance the equation: Zn(s) HCl(aq) ZnCl2(aq) H2(g).

b. Perform the calculation.

Stoichiometric mass-to-mass conversion If you were preparing

to carry out a chemical reaction in the laboratory, you would need to

know how much of each reactant to use in order to produce a certain

mass of product. This is one instance when you would use a mass-to-

mass conversion. In this calculation, you can find the mass of an

unknown substance in a chemical equation if you have the balanced

chemical equation and know the mass of one substance in the equation.

Example Problem 4 Stoichiometric Mass-to-Mass Conversion

How many grams of sodium hydroxide (NaOH) are needed to

completely react with 50.0 grams of sulfuric acid (H2SO4) to form

sodium sulfate (Na2SO4) and water?

Write the balanced equation.

2NaOH(aq) + H2SO4(aq) Na2SO4 + 2H2O(g)

Convert grams of sulfuric acid to moles NaOH.

50.0 g H2SO

4

1 mol H2SO

4

98.09 g H2SO

4

0.510 mol H2SO

4

0.510 mol H2SO

4

2 mol NaOH

1 mol H2SO

4

1.02 mol NaOH

Calculate the mass of NaOH needed.

1.02 mol NaOH 40.00 g NaOH

1 mol NaOH 40.8 g NaOH

50.0 grams of H2SO4 reacts completely with 40.8 grams of NaOH.



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Practice Problems

8. Balance each equation and solve the problem.

a. If 40.0 g of magnesium reacts with excess hydrochloric acid

(HCl), how many grams of magnesium chloride (MgCl2) are

produced?

Mg(s) HCl(aq) MgCl2(aq) H2(g)

b. Determine the mass of copper needed to react completely with a

solution containing 12.0 g of silver nitrate (AgNO3).

Cu(s) AgNO3(aq) Cu(NO3)2(aq) Ag(s)

c. How many grams of hydrogen chloride (HCl) are produced when

15.0 g of sodium chloride (NaCl) reacts with excess sulfuric acid

(H2SO4)?

NaCl(aq) H2SO4(aq) Na2SO4 HCl(g)

d. Calculate the mass of silver phosphate (Ag3PO4) produced if 30.0

g of silver acetate (AgCH3COO) reacts with excess sodium

phosphate (Na3PO4).

AgCH3COO(aq) Na3PO4(aq)

Ag3PO4(s) NaCH3COO(aq)

Steps in stoichiometric calculations Follow these basic steps

when performing stoichiometric calculations.

1. Write the balanced equation.

2. Determine the moles of the given substance using a

mass-to-mole conversion. Use the inverse of the molar

mass as the conversion factor.

3. Determine the moles of the unknown substance from

the moles of the given substance. Use the mole ratio

from the balanced equation as the conversion factor.

4. From the moles of the unknown substance, determine

the mass of the unknown substance using a mole-to-

mass conversion. Use the molar mass as the

conversion factor.



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11.3 Limiting Reactants

Rarely are the reactants in a chemical reaction present in the exact mole

ratios specified in the balanced equation. Usually, one or more of the

reactants are present in excess, and the reaction proceeds until all of one

reactant is used up. The reactant that is used up is called the limiting

reactant. The limiting reactant limits the reaction and, thus, determines

how much of the product forms. The left-over reactants are called

excess reactants.

How can you determine which reactant in a chemical reaction is

limited? First, find the number of moles of each reactant by multiplying

the given mass of each reactant by the inverse of the molar mass. Next,

determine whether the reactants are available in the mole ratio specified

in the balanced equation. A reactant that is available in an amount

smaller than that required by the mole ratio is a limiting reactant.

After the limiting reactant has been determined, calculate the amount

of product that can ideally form from the given amount of the limiting

reactant. To do this, multiply the given number of moles of the limiting

reactant by the mole ratio that relates the limiting reactant to the product.

Then, convert moles of product to mass using the molar mass of the

product as the conversion factor.

Example Problem 5 Determining the Limiting Reactant

In the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with

60.0 g of sulfuric acid (H2SO4).

2NaOH(aq) H2SO4(aq) Na2SO4 2H2O(g)

a. Which reactant is the limiting reactant?

b. What mass of Na2SO4 can be produced using the given quantities

of the reactants?

c. To determine the limiting reactant, calculate the actual ratio of

available moles of reactants.

40.0 g NaOH 1 mol NaOH

40.0 g NaOH 1.00 mol NaOH



9

60.0 g H2SO

4

1 mol H2SO

4

98.09 g H2SO

4

0.612 mol H2SO

4

So,

1.00 mol NaOH

0.612 mol H2SO

4

is available. Compare this ratio with the mole

ratio from the balanced equation:

2 mol NaOH

1 mol H2SO

4

, or

1 mol NaOH

0.5 mol H2SO

4

.

You can see that when 0.5 mol H2SO4 has reacted, all of the 1.00 mol of

NaOH would be used up. Some H2SO4 would remain unreacted. Thus,

NaOH is the limiting reactant.

d. To calculate the mass of Na2SO4 that can form from the given

reactants, multiply the number of moles of the limiting reactant

(NaOH) by the mole ratio of the product to the limiting reactant

and then multiply by the molar mass of the product.

1.00 mol NaOH 1 mol NaSO

4

2 mol NaOH

142.04 g Na2SO

4

1 mol Na2SO

4

71.0 g Na2SO4

71.0 g of Na2SO4 can form from the given amounts of the reactants.

Practice Problems

9. Ammonia (NH3) is one of the most common chemicals produced in

the United States. It is used to make fertilizer and other products.

Ammonia is produced by the following chemical reaction.

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

a. If you have 1.00103 g of N2 and 2.50103 g of H2, which is the

limiting reactant in the reaction?

b. How many grams of ammonia can be produced from the amount

of limiting reactant available?

c. Calculate the mass of excess reactant that remains after the

reaction is complete.



10

10. Aluminum reacts with chlorine to produce aluminum chloride.

a. Balance the equation: Al(s) Cl2(g) AlCl3(s).

b. If you begin with 3.2 g of aluminum and 5.4 g of chlorine, which

is the limiting reactant?

c. How many grams of aluminum chloride can be produced from

the amount of limiting reactant available?

d. Calculate the mass of excess reactant that remains after the

reaction is complete.

Reactions do not always continue until all of the reactants are used up.

Using an excess of the least expensive reactant in a reaction can ensure

that all of the more expensive reactant is used up, making the chemical

reaction more efficient and cost-effective. Using an excess of one

reactant can also speed up some reactions.

11.4 Percent Yield

Most chemical reactions do not produce the predicted amount of product.

Although your work so far with stoichiometric problems may have led

you to believe that chemical reactions proceed according to the balanced

equation and always produce the calculated amount of product, it’s not

true. Many reactions stop before all the reactants are used up, so less

product is formed than expected. Also, products other than those

expected sometimes form from competing chemical reactions, thereby

reducing the amount of the desired product.

The theoretical yield is the maximum amount of product that can be

produced from a given amount of reactant under ideal circumstances.

This is the amount you have been calculating in practice problems so far.

Chemical reactions hardly ever produce the theoretical yield. The actual

yield is the amount of product that is actually produced when a chemical

reaction is carried out in an experiment. It is determined by measuring

the mass of the product. Percent yield of product is the ratio of the

actual yield to the theoretical yield expressed as a percent.

Percent yield

actual yield (from an experiment)

theoretical yield (from stoichiometric calculations) 100



11

Percent yield tells you how efficient a chemical reaction is in producing

the desired product.

Example Problem 6 Calculating Percent Yield

Aspirin (C9H8O4) can be made from salicylic acid (C7H6O3) and acetic

anhydride (C4H6O3). Suppose you mix 13.2 g of salicylic acid with an

excess of acetic anhydride and obtain 5.9 g of aspirin and some water.

Calculate the percent yield of aspirin in this reaction.

Write the balanced equation.

2C7H6O3(s) C4H6O3(l) 2C9H8O4(s) + H2O(l)

Calculate the theoretical yield. Salicylic acid is the limiting reactant.

13.2 g C7H

6O

3

1 mol C7H

6O

3

138.1 g C7H

6O

3

0.0956 mol C7H

6O

3

0.0956 mol C7H

6O

3

2 mol C9H

8O

4

2 mol C7H

6O

3

0.0956 mol C9H

8O

4

0.0956 mol C9H

8O

4

180.2 g C9H

8O

4

1 mol C9H

8O

4

17.2 g C9H

8O

4

Calculate the percent yield.

5.9 g C9H

8O

4

17.2 g C9H

8O

4

100 34%

Practice Problems

11. Calculate the percent yield for each chemical reaction based on the

data provided.

a. theoretical yield: 25 g; actual yield: 20 g

b. theoretical yield: 55 g; actual yield: 42 g

c. theoretical yield: 5.2 g; actual yield: 4.9 g

12. Calculate the actual yield for each chemical reaction based on the

data provided.

a. theoretical yield: 20 g; percent yield: 95%

b. theoretical yield: 75 g; percent yield: 88%

c. theoretical yield: 9.2 g; percent yield: 62%



12

13. In an experiment, 10.0 g of magnesium reacted with excess

hydrochloric acid forming magnesium chloride.

Mg(s) 2HCl(aq) MgCl2(aq) H2(g)

At the completion of the reaction, 29.5 g of magnesium chloride

was produced. Calculate the theoretical yield and the percent yield.

Percent yield is important in the calculation of overall cost effectiveness

in industrial processes. Manufacturers must reduce the cost of making

products to the lowest level possible. For example, sulfuric acid (H2SO4)

is a raw material for many products, including fertilizers, detergents,

pigments, and textiles. The cost of sulfuric acid affects the cost of many

consumer items that use sulfuric acid as a raw material.

The manufacture of sulfuric acid is sometimes achieved using

a two-step process called the contact process. The two steps are

as follows.

S8(s) 8O2(g) 8SO2(g)

2SO2(g) O2(g) 2SO3(g)

The last step results in sulfuric acid as the product.

SO3(g) H2O(l) H2SO4(aq)

The first step produces almost 100 percent yield. The second step will

also produce a high yield if a catalyst is used. A catalyst is a substance

that speeds up a chemical reaction but is not used up in the chemical

reaction and does not appear in the chemical equation.

Chapter 11 Review

14. What is stoichiometry?

15. Write two questions that stoichiometry can help you answer about

the following chemical equation.

4HCl(aq) O2(g) 2H2O(l) 2Cl2(g)

16. Relate the law of conservation of mass to stoichiometry.

17. What is the difference between a limiting reactant and an excess

reactant?

Answer Key


13

Chapter 11

Practice Problems

1. a. 2H2O2(l) O2(g) 2H2O(l)

2 molecules H2O2 1 molecule O2 2 molecules H2O

2 moles H2O2 1 mole O2 2 moles H2O

68.04 g H2O2 32.00 g O2 36.04 g H2O, as shown below.

68.04 g reactant 68.04 g products

2 22 mol H O 2 2

2 2

34.02 g H O

1 mol H O 2 2

2

68.04 g H O

1 mol O

2

2

32.00 g O

1 mol O 2

2

32.00 g O

2 mol H O

2

2

18.02 g H O

1 mol H O 236.04 g H O

b. H2CO3(aq) H2O(l) CO2(g)

1 formula unit H2CO3 1 molecule H2O 1 molecule CO2

1 mole H2CO3 1 mole H2O 1 mole CO2

62.03 g H2CO3 18.02 g H2O 44.01 g CO2, as shown below.

62.03 g reactant 62.03 g products

2 31 mol H CO 2 3

2 3

62.03 g H CO

1 mol H CO 2 3

2

62.03 g H CO

1 mol H O

2

2

18.02 g H O

1 mol H O 2

2

18.02 g H O

1 mol CO

2

2

44.01 g CO

1 mol CO 244.01 g CO

Answer Key (continued)


14


4 molecules HCl 1 molecule O2

2 molecules H2O 2 molecules Cl2

4 mole HCl 1 mol O2 2 mole H2O 2 mole Cl2

146 g HCl 32.00 g O2 36.04 g H2O 142 g Cl2, as shown

below.

177.84 g reactants 177.84 g products

4 mol HCl36.46 g HCl

1 mol HCl

2

145.84 g HCl

1 mol O

2

2

32.00 g O

1 mol O 232.00 g O

22 mol H O 2

2

18.02 g H O

1 mol H O 2

2

36.04 g H O

2 mol Cl

2

2

70.90 g Cl

1 mol Cl 2141.80 g Cl

2. a. N2(g) O2(g) 2NO(g)

2 2 2 2

2 2

2 2

1 mol N 1 mol N 1 mol O 1 mol O; ; ; ;

1 mol O 2 mol NO 1 mol N 2 mol NO

2 mol NO 2 mol NO;

1 mol N 1 mol O



15

b. 4NH3(aq) 5O2(g) 4NO(g) 6H2O(l)

3 3 3 2

2 2 3

2 2

2 3 2

2 2

2 3

4 mol NH 4 mol NH 4 mol NH 5 mol O, , , ,

5 mol O 4 mol NO 6 mol H O 4 mol NH

5 mol O 5 mol O 4 mol NO 4 mol NO, , , ,

4 mol NO 6 mol H O 4 mol NH 5 mol O

6 mol H O 6 mol H O4 mol NO, ,

6 mol H O 4 mol NH 5 m

2

2

6 mol H O,

ol O 4 mol NO

c. 4HCl(aq) O2(g) 2H2O(l) 2Cl2

2

2 2 2

2 2 2 2

2 2 2

2 2

2

1 mol O4 mol HCl 4 mol HCl 4 mol HCl, , , ,

1 mol O 2 mol H O 2 mol Cl 4 mol HCl

1 mol O 1 mol O 2 mol H O 2 mol H O, , , ,

2 mol Cl 2 mol H O 4 mol HCl 1 mol O

2 mol H O 2 mol Cl 2 mol C, ,

2 mol Cl 4 mol HCl

2 2

2 2

l 2 mol Cl,

1 mol O 2 mol H O

3. 240 mol LiOH is needed.

4. 2KClO3(s) 2KCl(s) 3O2(g)

a. 15 mol O2

b. 3 mol KCl

c. 33 mol KClO3, or 30 mol KClO3 using significant figures

5. From the reaction 2Na(s) Cl2(g) 2NaCl(s), 321 g NaCl is

produced.

6. a. 2NaI(aq) Cl2(g) 2NaCl(aq) I2(s)

b. 213 g Cl2

7. a. Zn(s) 2HCl(aq) ZnCl2(aq) H2(g)

b. 365 g HCl

8. a. Mg(s) 2HCl(aq) MgCl2(aq) H2(g); 157 g MgCl2

b. Cu(s) 2AgNO3(aq) Cu(NO3)2(aq) 2Ag(s); 2.24 g Cu

c. 2NaCl(aq) H2SO4(aq) Na2SO4 2HCl(g); 9.36 g HCl

d. 3AgCH3COO(aq) Na3PO4(aq) Ag3PO4(s)

3NaCH3COO(aq); 25.1 g Ag3PO4

9. a. N2



16

b. 1.22103 g NH3

c. 34 g H2

10. a. 2Al(s) 3Cl2(g) 2AlCl3(s)

b. Cl2

c. 6.8 g AlCl3

d. 1.8 g Al

11. a. 80%

b. 76%

c. 94%

12. a. 19 g

b. 66 g

c. 5.7 g

13. theoretical yield: 39.2 g; percent yield: 75.3%

Chapter 11 Review

14. Stoichiometry is the study of quantitative relationships between

amounts of reactants used and products formed by a chemical

reaction.

15. Sample questions: What mass of HCl is needed to react completely

with a known mass of O2? How much water will be produced if a

given mass of HCl is used in the reaction?

16. The law of conservation of mass states that matter is neither created

nor destroyed; thus, in a chemical reaction, the mass of the reactants

must equal the mass of the products.

17. The limiting reactant limits the amount of product that can form

from the reaction. An excess reactant is left over after the reaction is

complete and all of the limiting reactant has been used up.

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