SECTION 9.

Limiting Reactants and

Percentage Yield
Teacher Notes and Answers
SECTION 3 Limiting Reactants and
Percentage Yield
1. oxygen gas; carbon
2. Because 0.400 mol of F​e3​ ​​O​4​were produced,
4 × 0.400 mol = 1.600 mol of ​H​2​were produced,
according to the mole ratio of F​e​3​​O​4​and ​H2​ ​.
The mass of 1.6 mol ​H​2​is 3.2 g ​H2​ ​because the
molar mass of hydrogen gas is about 2 g/mol.
3. In a real-world reaction, not all of the reactants
are used up. Reactants may also include
impurities or form byproducts through different
reactions.

Practice
Aa. ​H2​ ​​O​2​
Ab. 0.500 mol ​N2​ ​​H​4​
B. 79.7%

Review
1. The theoretical yield is the maximum amount
of product that can be produced from a given
amount of reactant. The actual yield is the
measured amount obtained.
2. oxygen gas
3. 93.8%
4. ​N2​ ​(g) + 3​H2​ ​(g) ⟶ 2N​H3​ ​(g); More ​H2​ ​is needed
because only 60.0% is converted to ammonia;
25.0 mol ​H​2​

Stoichiometry 1
SECTION 9.3

Limiting Reactants and

Percentage Yield
In the laboratory, a reaction is rarely carried out with Key Terms
­exactly the required amount of each of the reactants.
limiting reactant actual yield
­Usually, one or more of the reactants is present in excess: excess reactant percentage yield
there is more than the exact amount required to react. theoretical yield

Once one of the reactants is used up, the reaction stops.

No more product can be formed. The substance that is
completely used up first in a reaction is called the limiting
reactant. The limiting reactant is the reactant that limits the
amount of the other reactant that can combine and the
amount of product that can form in a chemical reaction.
A limiting reactant may also be referred to as a limiting
reagent. The substance that is not used up completely in a
reaction is called the excess reactant.
Consider the reaction between carbon and oxygen to
form carbon dioxide. According to the equation, one mole
of carbon reacts with one mole of oxygen to form one mole
of carbon ­dioxide.
C(s) + ​O2​ ​(g) → ​CO​2​(g)
Suppose you could mix 5 mol C with 10 mol O
​ ​2​. The
following diagram summarizes what would take place.

READING CHECK
5 carbon 10 oxygen 5 carbon 5 oxygen 1. If 15 mol C is mixed with
atoms molecules dioxide molecules 10 mol ​O2​ ​to form ​CO​2​, which
molecules in EXCESS
reactant is the limiting reactant?

There is more oxygen than is needed to react with the

carbon. Carbon is the limiting reactant in this situation, and
Which reactant would be the
it limits the amount of ​CO​2​that is formed. Oxygen is the excess reactant?
excess reactant, and 5 mol ​O2​ ​will be left over at the end of
the reaction.

2 CH A P TER 9
 SAMPLE PROBLEM
Silicon dioxide (quartz) is usually quite unreactive but reacts readily
with hydrogen fluoride according to the following equation.
Si​O2​ ​(s) + 4HF(g) → Si​F4​ ​(g) + 2​H2​ ​O(l)
If 6.0 mol HF is added to 4.5 mol Si​O​2​, which is the limiting reactant?

SOLUTION

1 ANALYZE Determine the information that is given and unknown.

Given: a mount of HF = 6.0 mol, amount of Si​O2​ ​= 4.5 mol
Unknown: limiting reactant

2 PLAN Determine how the unknown value can be calculated.

First, calculate the amount of a product that can be formed
with each of the given reactant amounts.
mol Si​F4​ ​
mol HF × ________
​ ​mol = mol Si​F4​ ​produced
mol HF
mol Si​F4​ ​
mol Si​O​2​× _________
​   ​= mol Si​F4​ ​produced
mol Si​O2​ ​

One amount will be less than the other. The lesser amount
represents the maximum amount of that product that can
possibly be formed. The limiting reactant is the reactant
that gives this lesser amount of product.

3 SOLVE Substitute the given values to determine the limiting reactant.

1 mol Si​F4​ ​
6.0 mol HF × __________
​ ​= 1.5 mol Si​F4​ ​produced
4 mol HF
1 mol Si​F4​ ​
4.5 mol Si​O​2​× __________
​   ​= 4.5 mol Si​F4​ ​produced
1 mol Si​O2​ ​

Under ideal conditions, the 6.0 mol HF present can make

1.5 mol Si​F4​ ​, and the 4.5 mol Si​O2​ ​present can make
4.5 mol Si​F​4​. Because 6.0 mol HF can make only 1.5 mol
Si​F4​ ​, HF is the limiting reactant.

4 CHECK Check the answer to see if it makes sense.

YOUR
The reaction requires four times the number of moles of
WORK
HF as it does moles of Si​O2​ ​. Because the amount of HF
available is less than four times the amount of Si​O2​ ​
available, HF is the limiting reactant.

Stoichiometry 3
PRACTICE

A. Some rocket engines use a mixture of hydrazine, N ​ 2​ ​​H​4​, and

hydrogen peroxide, ​H​2​​O​2​, as the propellant. The reaction is
given by the following equation.
​N​2​​H​4​(l) + 2​H2​ ​​O​2​(l) → N
​ 2​ ​(g) + 4​H2​ ​O(g)

a. Which is the limiting reactant in this reaction when

0.750 mol N​ 2​ ​​H​4​is mixed with 0.500 mol H
​ 2​ ​​O​2​?
b. How much of the excess reactant, in moles, remains unchanged?

a. What is given?
What is unknown?

What quantity of N
​ 2​ ​is produced by 0.750 mol N
​ 2​ ​​H​4​?

What quantity of N
​ 2​ ​is produced by 0.500 mol H
​ 2​ ​​O​2​?

Which reactant is the limiting reactant?

b. Usingthe amount of the limiting reactant calculated above, find
the amount of the excess reactant that is used in the reaction.

Subtract the amount of the excess reactant used in the r­ eaction

from the total amount of excess reactant. How much of the excess
reactant, in moles, remains unchanged?

Check your answers.

4 CH A P TER 9
 SAMPLE PROBLEM
Iron oxide, F​e​3​​O​4​, can be made in the laboratory by the reaction
between red-hot iron and steam according to the following equation.
3Fe(s) + 4​H​2​O(g) → F​e3​ ​​O​4​(s) + 4​H2​ ​(g)
a. When 36.0 g ​H​2​O is mixed with 67.0 g Fe, which reactant is the
limiting reactant?
b. What mass in grams of black iron oxide is produced?
c. What mass in grams of excess reactant remains when the
­reaction is completed?

SOLUTION

1 ANALYZE Determine the information that is given and unknown.

Given: mass of H
​ 2​ ​O = 36.0 g, mass of Fe = 67.0 g
Unknown: a . limiting reactant
b. mass of F​e3​ ​​O​4 produced
c. mass of excess reactant left over

2 PLAN Determine how the unknown values can be calculated.

a. The reactant yielding the smaller number of moles of
product is the limiting reactant.
molar mass factor mole ratio

mol Fe ​ mol F​e​3​​O​4​

g Fe × ​_______ × __________
​ ​ = mol F​e​3​​O​4​
g Fe mol Fe
molar mass factor mole ratio

mol H​ 2​ ​O mol F​e​3​​O​4​

g ​H2​ ​O × ​   _________  ​  ×  ​ __________  ​ = mol F​e​3​​O​4​
g ​H​2​O mol ​H2​ ​O

b. To find the maximum mass of F​e3​ ​​O​4​that can be ­produced,

use the amount of F​e3​ ​​O​4​in moles from the limiting reactant.
molar mass factor

g F​e ​3​​O​4​
mol F​e3​ ​​O​4​from limiting reactant × __________
​ ​= g F​e3​ ​​O​4​
mol F​e3​ ​​O​4​

Stoichiometry 5
 c. Use the moles of product found using the limiting reactant to
determine the amount of the excess reactant that is consumed.
Then subtract the amount consumed from the original amount.
mol excess reactant g excess reactant
  ​× ​_________________
mol product × ​_________________
       ​
mol product mol excess reactant
= grams of excess reactant consumed
original mass − mass consumed = mass of reactant remaining

3 SOLVE Substitute the given values to determine the unknown values.

a. The molar masses are 18.02 g/mol H​2​O, 55.85 g/mol Fe, and
231.55 g/mol F​e​3​​O​4.
1 mol F​e2​ ​​O​4​
1 mol Fe ​× ___
67.0 g Fe × ​__ ​   ​= 0.400 mol F​e​3​​O​4​
55.85 g Fe 3 mol Fe

1 mol H
​ ​2​O 1 mol F​e3​ ​​O​4​
36.0 g ​H2​ ​O ___
​     ​× ___
​     ​= 0.499 mol F
​ e​3​​O​4​
18.02 g ​H2​ ​O 4 mol ​H2​ ​O
Fe is the limiting reactant because the given amount of
Fe can make only 0.400 mol ​Fe​3​​O​4​, which is less than the
0.499 mol F ​ e​3​​O​4​that the given amount of H
​ 2​ ​O would produce.
231.55 g F​e​3​​O​4​
b. 0.400 mol F​e3​ ​​O​4​× ___
​   
  ​= 92.6 g F​e3​ ​​O​4​
1 mol F​e​3​​O​4​

​ ​2​O
4 mol H 18.02 g ​H2​ ​O
c. 0.400 mol F​e3​ ​​O​4​× ___
​   ​× ___​    ​
1 mol F​e​3​​O​4​ 1 mol ​H2​ ​O
= 28.8 g ​H2​ ​O consumed
36.0 g ​H2​ ​O − 28.8 g ​H2​ ​O consumed = 7.2 g H
​ 2​ ​O remaining

4 CHECK Check the answer to see if it makes sense.

YOUR
The mass of original reactants is 67.0 g + 36.0 g = 103.0 g. The
WORK
mass of F​e3​ ​​O​4​+ unreacted water is 92.6 g + 7.2 g = 99.8 g. The
difference of 3.2 g is the mass of hydrogen that is produced with
the F​e​3​​O​4​. Therefore, the answers make sense in the context of
the conservation of mass.

Critical Thinking
2. Evaluate How could you check the statement in Step 4 that
the 3.2 g difference is the mass of hydrogen that is ­produced?

6 CH A P TER 9
 Comparing the actual and theoretical yields help
chemists determine the reaction’s efficiency.

The product amounts that you calculate in ideal stoichiometry

problems give you the theoretical yields for a reaction. The
theoretical yield is the maximum amount of the product that
can be produced from a given amount of reactant. In most
chemical reactions, the amount of product obtained is less
than the theoretical yield.
There are many reasons that the actual yield is usually less
than the theoretical yield. Reactants may contain impurities or
may form byproducts in competing side reactions. Also, many
reactions do not go to completion. As a result, less product is
produced than ideal stoichiometric calculations predict. The
actual yield of a product is the measured amount of that
product obtained from a reaction.
Chemists are usually interested in the efficiency of a
reaction. The efficiency is expressed by comparing the actual
and ­theoretical yields. The percentage yield is the ratio of the
actual yield to the theoretical yield, multiplied by 100.
actual yield
percentage yield = ___
​     ​× 100
theoretical yield

theoretical yield actual yield

10 oxygen 5 carbon
molecules atoms
Consider the synthesis of carbon
dioxide from carbon and oxygen as
discussed earlier in this section.
In theory, 10 oxygen molecules and
5 carbon atoms can combine to
form 5 carbon dioxide molecules.
In practice, a reaction with an 80%
5 carbon dioxide 4 carbon dioxide percentage yield will only
molecules molecules produce 4 carbon dioxide molecules.

READING CHECK

3. Give two reasons that the actual yield of a reaction might

be less than the theoretical yield of a reaction.

Stoichiometry 7
 SAMPLE PROBLEM
One industrial method of preparing chlorobenzene, ​C6​ ​​H​5​Cl, is to
react benzene, ​C​6​​H​6​, with chlorine, as represented by the following
equation.
​​C​6​H​6​(l) + ​Cl​2​(g) → C
​ 6​ ​​H​5​Cl(l) + HCl(g)
When 36.8 g ​C​6​​H​6​react with an excess of C ​ l​2​, the actual yield of
​C6​ ​​H​5​Cl is 38.8 g. What is the percentage yield of C ​ 6​ ​​H​5​Cl?

SOLUTION

1 ANALYZE Determine the information that is given and unknown.

Given: mass of C
​ 6​ ​​H​6​= 36.8 g, mass of ​Cl​2​= excess
actual yield of C ​ 6​ ​​H​5​Cl = 38.8 g
Unknown: percentage yield of ​C​6​​H​5​Cl

2 PLAN Determine how the unknown value can be calculated.

First do a mass-mass calculation to find the theoretical yield of
​C6​ ​​H​5​Cl. Then the percentage yield can be found.

molar mass factor molar mass factor

mol C mol C
​ 6​ ​​H​6​ ___ ​ ​6​​H​5​Cl ___
g ​C​6​​H​5​Cl
g ​C​6​​H​6​× __
​ ​× ​    ​× ​ ​= g ​C6​ ​H5​ ​Cl
g ​C6​ ​​H​6​ mol ​C6​ ​​H​6​ mol ​C6​ ​​H​5​Cl

mole ratio

actual yield
percentage yield ​C​6​H5​ ​Cl = ___
​     ​× 100
theoretical yield

3 SOLVE Substitute the given values to determine the unknown values.

The molar masses are 78.12 g/mol C​6​H​6 and 112.56 g/mol C​6​H5​ ​Cl.
mol ​C​6​​H​6​ 112.56 g ​C6​ ​​H​5​Cl
1 mol ​C6​ ​​H​5​Cl ___
36.8 g ​C​6​​H​6​× ___
​   ​ × ___ ​   ​× ​
       ​
78.12 g ​C6​ ​​H​6​ 1 mol ​C6​ ​​H​6​ mol ​C6​ ​​H​5​Cl

= 53.0 g ​C6​ ​​H​5​Cl (theoretical yield)

38.8 g
percentage yield = ​__​× 100 = 73.2%
53.0 g

8 CH A P TER 9
PRACTICE

B. Methanol
can be produced through the reaction of
CO and ​H2​ ​in the presence of a catalyst.
CO(g) + ​2H​2​(g) → ​CH​3​OH(l)
If 75.0 g of CO react to produce 68.4 g C
​ H​3​OH, what
​ H​3​OH?
is the percentage yield of C
What is known?
What is unknown?

Write the mass-mass calculation needed to find the

theoretical yield of ​CH​3​OH.

Use the periodic table to find the molar mass of CO

and the molar mass of ​CH​3​OH.

Compute the theoretical yield.

Use the actual yield and the theoretical yield to compute

the percentage yield.

Check your answer.

Stoichiometry 9
 SECTION 9.3 REVIEW
VOCABULARY
1. Describe the difference between actual yield and theoretical yield.

REVIEW
2. Carbon disulfide burns in oxygen to yield carbon dioxide and sulfur dioxide
according to the following chemical equation.

​ CS​2​(l) + ​3O​2​(g) → ​CO​2​(g) + ​2SO​2​(g)

If 1.00 mol ​CS​2​is combined with 1.00 mol O

​ 2​ ​, identify the limiting reactant.

3. Quicklime, CaO, can be prepared by roasting limestone, C

​ aCO​3​, according
to the following reaction.

​CaCO​3​(s) → CaO(s) + CO​2​(g)

When 2.00 × ​10​3​g ​CaCO​3​are heated, the actual yield of CaO is

1.05 × ​10​3​g. What is the percentage yield?

Critical Thinking
4. ANALYZING DATA A chemical engineer calculated that 15.0 mol ​H2​ ​were
needed to react with excess ​N2​ ​to prepare 10.0 mol ​NH​3​. But the actual yield
is 60.0%. Write a balanced chemical equation for the reaction. Is the amount
of ​H​2​needed to make 10.0 mol N ​ H​3​more, the same, or less than 15 mol?
How many moles of H ​ ​2​are needed?

10 CH A P TER 9
 Math Tutor Using Mole Ratios

The coefficients in a balanced equation represent the relative amounts in moles of

reactants and products. You can use the coefficients of two of the substances in the
equation to set up a mole ratio. A mole ratio is a conversion factor that relates the
amounts in moles of any two substances involved in a chemical reaction.

Problem-Solving TIPS
• When solving stoichiometric problems, always start with a balanced
chemical ­equation.
• Identify the amount known from the problem (in moles or mass).
• If you are given the mass of a substance, use the molar mass factor as a conversion
factor to find the amount in moles. If you are given the amount in moles of a
­substance, use the molar mass factor as a conversion factor to find the mass.

SAMPLE

If 3.61 g of aluminum react completely with excess CuC​l​2​, what mass

of copper metal is produced? Use the balanced equation below.
2Al(s) + 3CuC​l2​ ​(aq) → 2AlC​l3​ ​(aq) + 3Cu(s)
First, convert the amount of aluminum from a mass in grams to a
number of moles by using the molar mass of aluminum.

1 mol Al  ​= 0.134 mol Al

3.61 g Al = 3.61 g Al × ​ __
26.98 g Al

Next, apply the mole ratio of aluminum to copper to find the moles
of copper ­produced.

​ 3 mol Cu ​= mol Cu

mol Al × __
2 mol Al

3 mol Cu ​= 0.201 mol Cu

0.134 mol Al = 0.134 mol Al × ​ __
2 mol Al

Then, convert moles of copper to mass of copper in grams by using

the molar mass of copper.

63.55 g Cu
0.201 mol Cu = 0.201 mol Cu × ​ __ ​= 12.8 g Cu
1 mol Cu

Practice Problems: Chapter Review practice problems 8 and 9

Stoichiometry 11