Chapter 6:
Quantities in
Chemical
Reactions
6-1
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Questions for Consideration
1.
2.
3.
4.
5.
6.
7.
What do the coefficients in balanced equations
represent?
How can we use a balanced equation to relate the
number of moles of reactants and products in a chemical
reaction?
How can we use a balanced equation to relate the mass
of reactants and products in a chemical reaction?
How do we determine which reactant limits the amount
of product that can form?
How can we compare the amount of product we actually
obtain to the amount we expect to obtain?
How can we describe and measure energy changes?
How are heat changes involved in chemical reactions?
6-2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Chapter 6 Topics:
1.
2.
3.
4.
5.
6.
7.
The Meaning of a Balanced Equation
Mole-Mole Conversions
Mass-Mass Conversions
Limiting Reactants
Percent Yield
Energy Changes
Heat Changes in Chemical Reactions
6-3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Introduction


How can we predict
amounts of reactants
and products in a
reaction, such as that in
an internal combustion
engine?
How can we predict the
amount of heat
generated or absorbed
during a reaction?
Figure 6.2
6-4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
6.1 The Meaning of a Balanced
Equation

What does a balanced equation tell us about the
relative amounts of reactants and products? Let’s
consider the combustion of propane:
Figure 6.4
6-5
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
What do the coefficients in a balanced
chemical equation mean?

Balanced, the equation is:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

The coefficients in the balanced equation tell us
about the relative numbers of reactants that combine
and products that form. There is an implied
coefficient of 1 in front of C3H8(g).
6-6
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
What do the coefficients in a balanced
chemical equation mean?
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

For every 1 molecule of propane that reacts with 5
molecules of oxygen gas, 3 molecules of carbon
dioxide and 4 molecules of water are produced.
Figure 6.4
6-7
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
What do the coefficients in a balanced
chemical equation mean?
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)


The coefficients in the balanced equation also tell us
about the relative numbers of moles of reactants and
products.
For every 1 mole of C3H8(g) that reacts with 5 moles
of O2(g), 3 moles of CO2(g) and 4 moles of H2O(g) are
formed.
6-8
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Summary – The Meaning of the Coefficients
6-9
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Amounts of
Reactants and Products
 The process of determining the amount of a reactant
or product from another reactant or product in a
reaction is called stoichiometry.
 We use the mole relationships provided in a
balanced equation to calculate amounts of reactants
and products in a reactions.
6-10
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
6.2 Mole-Mole Conversions
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)



A mole ratio is used to relate the number of
moles of one reactant or product to another.
Mole ratios are obtained from the coefficients in
the balanced equation.
For example, the mole ratio of O2 to C3H8 is 5:1
or:
5 mol O 2
1 mol C3 H 8
6-11
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Mole Ratios
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)


The mole ratio of O2 to C3H8 allows us to
calculate the amount of O2 that will react with
any amount of C3H8 that reacts. The mole ratio
is used as a conversion factor in a dimensional
analysis equation.
If 0.40 mol C3H8 reacts:
5 mol O2
0.40 mol C3H8 
= 2.0 mol O2
1 mol C3H8
6-12
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Mole-Mole Conversions

Benzene (C6H6) burns in air according to the
following equation:
2C6H6(g) + 15O2(g)  12CO2(g) + 6H2O(g)
1.
2.
3.
What is the mole ratio of O2 to C6H6?
How many moles of O2 are required to react
with each mole C6H6?
How many moles of O2 are required to react
with 0.38 mole of C6H6?
6-13
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Mole-Mole Conversions
2C6H6(g) + 15O2(g)  12CO2(g) + 6H2O(g)
1. What is the mole ratio of O2 to C6H6?
Based on the coefficients in the balanced chemical equation,
the mole ratio can be written two ways:
15 moles O2
2 moles C6H6
OR
2 moles C6H6
15 moles O2
How many moles of O2 are required to react with each
mole C6H6?
15/2 or 7.5 moles of O2 are required to react with each mole
of C6H6
3. How many moles of O2 are required to react with 0.38
mole of C6H6?
2.
0.38 mol C6 H 6
15 mol O2

= 2.9 mol O2
2 mol C6 H 6
6-14
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Mole Ratios
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)


What is the mole ratio for determining the moles
of CO2 that will be produced when 2.3 mol O2
reacts?
How many moles of CO2 will be produced?
3 mol CO 2
2.3 mol O 2 
= ? mol CO 2
5 mol O 2
moles CO2 = 2.3 mol O2
3 mol CO2

= 1.4 mol CO2
5 mol O2
6-15
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Mole Ratios
2Al(s) + 3Cl2(g)  2AlCl3(g)

How many moles of Cl2 are required to prepare
0.62 mol AlCl3?
3 mol Cl2
moles Cl2 = 0.62 mol AlCl3 
= 0.93 mol Cl2
2 mol AlCl3
6-16
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
6.3 Mass-Mass Conversions


2Na(s) + Cl2(g)  2NaCl(s)
What mass of chlorine gas is required to
react with 9.20 grams of sodium?



We don’t measure reactants and products in
moles, but we commonly measure their mass.
The balanced equation does not tell us a mass
relationship.
How do we convert grams of reactant or
product to moles?
6-17
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Mass-Mass Conversions
2Na(s) + Cl2(g) 
2NaCl(s)
(9.20 g Na)
1. We convert grams to moles using molar
mass:
For a
review, see
Section 4.2
or Math
Toolbox 4.1.
Figure from p. 220
1 mol Na
9.20 g Na 
 0.400 mol Na
22.99 g Na
6-18
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Mass-Mass Conversions
2Na(s) +
Cl2(g) 
2NaCl(s)
(9.20 g Na)
(0.400 mol)
2. Next we relate moles of Na to moles of Cl2
using the mole ratio:
Figure from p. 220
1 mol Cl2
0.400 mol Na 
 0.200 mol Cl2
2 mol Na
6-19
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Mass-Mass Conversions
Cl2(g) 
2Na(s) +
(9.20 g Na)
(0.400 mol Na)
2NaCl(s)
(0.200 mol Cl2)
3. The last step is to convert moles of Cl2 to
grams using the molar mass of Cl2:
Figure from p. 220
0.200 mol Cl2
70.90 g Cl2

 14.2 g Cl2
1 mol Cl2
6-20
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Mass-Mass Conversions
2Na(s) +
Cl2(g) 
2NaCl(s)
(9.20 g Na)
(14.2 g Cl2)
(0.400 mol Na) (0.200 mol Cl2)
3. The last step is to convert moles of Cl2 to
grams using the molar mass of Cl2:
0.200 mol Cl2
70.90 g Cl2

 14.2 g Cl2
1 mol Cl2
6-21
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Steps for Mass-Mass Conversions
From page 221
1 mol Na 1 mol Cl2 70.90 g Cl2
9.20 g Na 


 14.2 g Cl2
22.99 g Na 2 mol Na 1 mol Cl2
6-22
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Mass-Mass Conversions

2Na(s) + Cl2(g)  2NaCl(s)
What mass of NaCl should be produced when
9.20 g Na reacts with 14.2 g Cl2?
1 mol Na
2 mol NaCl 58.44 g NaCl
9.20 g Na 


 23.4 g NaCl
22.99 g Na
2 mol Na
1 mol NaCl
14.2 g Cl2 
1 mol Cl2
70.90 g Cl2

2 mol NaCl 58.44 g NaCl

 23.4 g NaCl
1 mol Cl2
1 mol NaCl
Or, from the law of conservation of mass:
9.20 g Na + 14.2 g Cl 2 = 23.4 g NaCl
6-23
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Mass-Mass Conversions
2Na(s) + Cl2(g)  2NaCl(s)

What mass of NaCl should be produced when
9.20 g Na reacts with 14.2 g Cl2?
Figure 6.6
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
6-24
Activity: Mass-Mass Conversions

Given:
2 H2(g) + O2(g)  2 H2O(g)
If 1.8 × 108 g of hydrogen was used in the liftoff
during a shuttle launch, then:
1. What mass of oxygen gas was consumed?
2. What mass of water vapor was produced?
3. Does the mass of the water vapor equal the
masses of the reactants? If so, then what law
describes this observation?
6-25
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Mass-Mass Conversions
1.
2 H2(g) + O2(g)  2 H2O(g)
What mass of oxygen gas was consumed?
Start with the number that was given in the problem –
1.8×108 g of H2(g). Convert to moles via the MM(H2), then
use a mole ratio to relate H2 to O2. Finally, convert from
moles of O2 to grams of O2 using the MM(O2).
1 mol H 2
1 mol O 2
32.00 g O 2
1.8×10 g H 2 ×
×
×
= 1.4×109 g O 2
2.016 g H 2 2 mol H 2 1 mol O 2
8
Mass H2 MM(H2)
(Given in
Problem)
Moles H2
Mole
Ratio
MM(O2)
Moles O2
Mass O2
6-26
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Mass-Mass Conversions
2.
2 H2(g) + O2(g) → 2 H2O(g)
What mass of water vapor was produced? Start again
with the number that was given in the problem: 1.8×108 of
H2(g). Convert to moles using the MM(H2), then use a mole
ratio to relate H2 to H2O. Finally, convert from moles of H2O
to grams of H2O using the MM(H2O).
8
1.8×10 g H 2 ×
1 mol H 2
2.016 g H 2
Mass H2 MM(H2)
(Given in
Problem)
×
2 mol H 2O
Moles H2
2 mol H 2
18.02 g H 2O
×
= 1.6×109 g H 2O
1 mol H 2O
Mole
Ratio
MM(H2O)
Moles
H2O
Mass H2O
6-27
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Mass-Mass Conversions

Given:
2 H2(g) +
O2(g)  2 H2O(g)
1.8 × 108 g H2 + 1.4 × 109 g O2 = 1.6 × 109 g H2O
3. Does the mass of the water vapor equal the masses of
the reactants?
Yes or close to it – hydrogen and oxygen added
together equals 1.58 × 109 g H2O.
If so, then what law describes this observation?
The Law of Conservation of Mass
6-28
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
The Law of Conservation of Mass
Figure 6.6

The law of conservation of mass states that the masses
of the reactants that are consumed must equal the
masses of the products that are formed.
Mass consumed reactants = Mass formed products
6-29
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Mass-Mass Conversions

When aluminum metal is exposed to oxygen gas, a
coating of aluminum oxide forms on the surface of
the aluminum. The balanced equation for the
reaction of aluminum metal with oxygen gas is:
4Al(s) + 3O2(g)  2Al2O3(s)
Suppose a sheet of pure aluminum gains 0.0900 g of
mass when exposed to air. Assume that this gain can
be attributed to its reaction with oxygen.
1.
2.
3.
What mass of O2 reacted with the Al?
What mass of Al is reacted?
What mass of Al2O3 is formed?
6-30
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Mass-Mass
Conversions
4Al(s) + 3O2(g)  2Al2O3(s)
Suppose a sheet of pure aluminum gains 0.0900 g of mass
when exposed to air. Assume that this gain can be
attributed to its reaction with oxygen.
1.
What mass of O2 reacted with the Al?
0.0900 g O2
2.
What mass of Al is reacted?
1 mol O 2
4 mol Al 26.98 g Al
0.0900 g O 2 ×
×
×
= 0.101 g Al
32.00 g O 2 3 mol O 2 1 mol Al
3.
What mass of Al2O3 is formed?
0.0900 g O 2 ×
1 mol O 2
32.00 g O 2
×
2 mol Al 2O 3
3 mol O 2
101.96 g Al 2O 3
×
= 0.191 g Al 2O 3
1 mol Al
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
6-31
6.4 Limiting Reactants



When two reactants are mixed, one usually does not react
completely because there is too much of it.
In this reaction solid magnesium metal reacts with aqueous
hydrochloric acid (A). One reactant is in excess in B and the
other reactant is in excess in C.
Can you identify the excess reactant in each case?
Figure 6.7
6-32
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Limiting Reactant


The limiting reactant is the reactant that reacts
completely and is therefore not present when the
reaction is complete.
Since the limiting reactant reacts completely, its
amount determines the amount of product that
can form.
6-33
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
What is the limiting reactant?
Cu(s) + AgNO3(aq) 
Figure from
Example 6.3
6-34
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Consider an Analogy:
The construction of a model solar car
Figure 6.8 (A)
6-35
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
If you have 4 frames, 5 solar cells, 6
motors, and 12 wheels, how many solar
cars can you make?
Figure 6.8 (A and B)
6-36
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
What is the limiting part?
Figure 6.8 (B)


The wheel is the limiting part because it will be
completely used up.
The number of cars made depends on the number
of wheels available. If there were 4 more wheels,
then another car could have been made with the
remaining parts.
6-37
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
How many solar cars?
Figure 6.8 B( and C)
6-38
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Limiting Reactants at a Molecular
Level

In a chemical reaction, the balanced equation
tells us the relative number of molecules (or
moles) that combine in the reaction.
Figure 6.9
6-39
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Limiting Reactants at a Molecular
Level

If reactants are not present in this ratio, then
there will be a limiting reactant and excess of
the other reactant.
Figure 6.10
6-40
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Limiting Reactant
2H2(g) + O2(g)  2H2O(g)


Complete the after picture. How many H2O
molecules form?
What is the limiting reactant? What is in excess?
Figure from p. 251
6-41
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solution: Limiting Reactant
2H2(g) + O2(g)  2H2O(g)
What is the limiting reactant?
Using molecule ratios from the balanced chemical
equation:
2 molecules H 2 O
3 molecules O2 
= 6 molecules H 2 O
1 molecule O2
2 molecules H 2 O
8 molecules H 2 
= 8 molecules H 2 O
2 molecules H 2
The limiting reactant is O2 because there is only enough of
it to create 6 molecules of H2O.
Which reactant is left over?
H2 is left over at the end of the reaction.
6-42
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
2H2(g) + O2(g)  2H2O(g)


You can also use “molecule” ratios to determine the limiting
reactant. The 8 H2 molecules need 4 O2 molecules to react
with them. There are only 3 O2 molecules, so all the H2 cannot
react.
H2 is in excess, and O2 is the limiting reactant.
Figure from p. 251
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
6-43
Activity Solution: Limiting Reactant

The molecular-level diagram shows a mixture of reactant
molecules (three O2 molecules and 8 H2 molecules) for the
following reaction:
2H2(g) + O2(g)  2H2O(g)
The after picture should have 6 H2O molecules and 2 H2
molecules:
Figure from
p. 251
6-44
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Steps for Determining the
Limiting Reactant
1.
2.
Calculate the amount of one reactant (B) needed to
react with the other reactant (A).
Compare the calculated amount of B (amount
needed) to the actual amount of B that is given.
a.
If calculated B = actual B, there is no limiting reactant.
Both A and B will react completely.
b.
If calculated B > actual B, B is the limiting reactant. Only
B will react completely.
a.
If calculated B < actual B, A is the limiting reactant. Only
A will react completely.
6-45
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Limiting Reactants
(Mole Scale)

The balanced equation for the reaction of
phosphorus and oxygen gas to form diphosphorus
pentoxide is:
P4(s) + 5O2(g)  2P2O5(s)
What is the limiting reactant when each of the
following sets of quantities of reactants is mixed?
a.
b.
c.
0.50 mol P4 and 5.0 mol O2
0.20 mol P4 and 1.0 mol O2
0.25 mol P4 and 0.75 mol O2
6-46
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Limiting Reactants
(Mole Scale)
P4(s) + 5O2(g) → 2P2O5(s)
What is the limiting reactant when each of the
following sets of quantities of reactants is mixed?
a.
0.50 mol P4 and 5.0 mol O2
Each mole of P4 requires 5 moles of O2, so 0.50
mol P4 requires:
5 mol O2
0.50 mol P4 
= 2.5 mol O2
1 mol P4
The amount of O2 present is more than the amount
required, so P4 is the limiting reactant and O2 is
present in excess.
6-47
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Limiting Reactants
(Mole Scale)
P4(s) + 5O2(g)  2P2O5(s)
What is the limiting reactant when each of the
following sets of quantities of reactants is mixed?
b.
0.20 mol P4 and 1.0 mol O2
Each mole of P4 requires 5 moles of O2, so 0.20
mol P4 requires:
5 mol O2
0.20 mol P4 
= 1.0 mol O2
1 mol P4
Since this is the amount of O2 present, there is no
limiting reactant. Both reactants are consumed
completely.
6-48
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Limiting Reactants
(Mole Scale)
P4(s) + 5O2(g)  2P2O5(s)
What is the limiting reactant when each of the
following sets of quantities of reactants is mixed?
c.
0.25 mol P4 and 0.75 mol O2
Each mole of P4 requires 5 moles of O2, so 0.25 mol
P4 requires:
5 mol O2
0.25 mol P4 
= 1.25 mol O2
1 mol P4
Since the required amount of O2 is greater than that
present, the limiting reactant is O2 and P4 is present
in excess.
6-49
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Limiting Reactants – Mole Scale

1.
2.
N2(g) + 3H2(g)  2NH3(g)
Identify the limiting reactant when the following
are mixed:
a) 2.0 mol N2 and 5.0 mole H2
b) 3.10 mole N2 and 10.2 mol H2
How many moles of NH3 can be produced in
each case?
6-50
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Limiting Reactants – Mole Scale

1.
2.
N2(g) + 3H2(g)  2NH3(g)
Identify the limiting reactant when the following
are mixed:
a) 2.0 mol N2 and 5.0 mole H2
b) 3.10 mole N2 and 10.2 mol H2
How many moles of NH3 can be produced in
each case?
a) 3.3 mole NH3
b) 6.2 mole NH3
6-51
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Limiting Reactants (Mass Scale)



Two approaches can be used when masses of the
reactants are given. In each approach, it is
necessary to convert masses of reactants to moles
of reactants as a first step.
In the preferred approach, determine the limiting
reactant on a mole scale as before, then proceed to
calculate the mass.
In an alternate approach, calculate the amount of
product predicted from each reactant. The
calculation that gives the least amount of product
identifies the limiting reactant.
6-52
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Limiting Reactants
(Mass Scale)

The balanced chemical equation for the reaction of
aluminum metal and chlorine gas is:
2Al(s) + 3Cl2(g)  2AlCl3(s)
Assume 0.40 g Al is mixed with 0.60 g Cl2.
a. What is the limiting reactant?
b. What is the maximum amount of AlCl3, in grams,
that can be produced?
6-53
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Limiting Reactants
(Mass Scale)
2Al(s) + 3Cl2(g)  2AlCl3(s)
Assume 0.40 g Al is mixed with 0.60 g Cl2.
a.
What is the limiting reactant?
Calculate the moles of each reactant and determine which
is the limiting reactant as in earlier examples:
0.40 g Al ×
1 mol Al
= 0.148 mol Al
26.98 g Al
0.60 g Cl 2 ×
1 mol Cl 2
= 0.00846 mol Cl 2
70.90 g Cl 2
For every 2 moles of Al, 3 moles of Cl2 are required, so
0.148 moles of Al requires:
0.148 mol Al ×
3 mol Cl 2
= 0.222 mol O2
2 mol Al
The limiting reactant is Cl2 because the required amount
is greater than the amount present.
6-54
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Limiting Reactants
(Mass Scale)
2Al(s) + 3Cl2(g)  2AlCl3(s)
Assume 0.40 g Al is mixed with 0.60 g Cl2.
b. What is the maximum amount of AlCl3, in grams,
that can be produced?
All of the 0.00846 mol Cl2 present in the system
will react, so we use this amount to calculate the
mass of product:
0.00846 mol Cl 2 ×
2 mol AlCl 3
3 mol Cl 2
133.3 g AlCl 3
×
= 0.75 g AlCl 3
1 mol AlCl 3
6-55
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
6.5 Percent Yield


The amount of product
actually obtained in the
lab (actual yield) is
usually less than the
amount predicted by
calculations (theoretical
yield).
Yields describe the
amount of product, and
can be in mass units,
moles, or number of
molecules.
Figure 6.11
6-56
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Percent Yield

The percent yield describes how much of a
product is actually obtained relative to the
amount that should form assuming
complete reaction of the limiting reactant.
actual yield
Percent Yield =
100%
theoretical yield
6-57
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Percent Yield

2Na(s) + Cl2(g)  2NaCl(s)

If 0.20 mol chlorine reacts with excess sodium,
and 0.40 mol NaCl are produced, what is the
percent yield for the reaction?
Theoretical yield = 0.20 mol Cl 2 ×
2 mol NaCl
= 0.40 mol NaCl
1 mol Cl 2
actual yield
Percent Yield =
× 100%
theoretical yield
0.40 mol
Percent Yield =
× 100% = 100%
0.40 mol
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
6-58
Activity: Percent Yield

2Na(s) + Cl2(g)  2NaCl(s)

If 0.450 mol chlorine reacts with excess
sodium, and 0.385 mol NaCl are produced,
what is the percent yield for the reaction?
2 mol NaCl
Theoretical yield = 0.450 mol Cl 2 ×
= 0.900 mol NaCl
1 mol Cl 2
actual yield
Percent Yield =
× 100%
theoretical yield
0.385 mol
Percent Yield =
× 100% = 42.8%
0.900 mol
6-59
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity: Percent Yield

A student was synthesizing aspirin in the laboratory. Using
the amount of limiting reactant, she calculated the amount of
aspirin that should form as 8.95 g. When she weighed her
aspirin product on the balance, its mass was 7.44 g.
a.
What is the actual yield of the aspirin?
b.
What is the theoretical yield of the aspirin?
c.
Calculate the percent yield for this synthesis.
6-60
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Activity Solutions: Percent Yield

A student was synthesizing aspirin in the laboratory. Using
the amount of limiting reactant, she calculated the amount of
aspirin that should form as 8.95 g. When she weighed her
aspirin product on the balance, its mass was 7.44 g.
a.
What is the actual yield of the aspirin?
7.44 g aspirin
b.
What is the theoretical yield of the aspirin?
8.95 g aspirin
c.
Calculate the percent yield for this synthesis.
actual yield
Percent Yield =
× 100%
theoretical yield
7.44 g
=
× 100% = 83.1%
8.95 g
6-61
Copyright © McGraw-Hill Education. Permission required for reproduction or display.