Chapter 6:
Quantities in
Chemical
Reactions
Part 3
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6-1
Limiting Reactants – Mole Scale

N2(g) + 3H2(g)  2NH3(g)
1.
Identify the limiting reactant when the
following are mixed:
(a) 2.0 mol N2 and 5.0 mole H2
6-2
Limiting Reactants – Mole Scale

1.
a)
N2(g) + 3H2(g)  2NH3(g)
Identify the limiting reactant when the following are mixed:
2.0 mol N2 and 5.0 mole H2
If 2.0 mol N2 is reacted, the needed H2 mol:
H 2 mole
Needed H 2 mol  reacted N 2 mol 
N 2 mole
3 mol H 2
 2.0 mol N 2 
1 mol N 2
 6.0 mol H 2  actual H 2 mole
Hence, H 2 is not enough. It is limiting reatant.
6-3
Limiting Reactants – Mole Scale

N2(g) + 3H2(g)  2NH3(g)
1.
Identify the limiting reactant when the
following are mixed:
(b) 3.10 mole N2 and 10.2 mol H2
6-4
Limiting Reactants – Mole Scale


N2(g) + 3H2(g)
1.
Identify the limiting reactant when the
following are mixed:
a)
b)
2.
2NH3(g)
2.0 mol N2 and 5.0 mole H2 ( Have found that
H2 is limiting)
3.10 mole N2 and 10.2 mol H2
How many moles of NH3 can be produced
in each case?
6-5
Limiting Reactants – Mole Scale

1.
a)
b)
2.
N2(g) + 3H2(g) 2NH3(g)
Identify the limiting reactant when the following are mixed:
2.0 mol N2 and 5.0 mole H2 ( Have found that H2 is limiting)
3.10 mole N2 and 10.2 mol H2
How many moles of NH3 can be produced in each case?
The amount of product depends on the amount of
limiting reactant.
NH3mole
NH3 mol  reacted H 2 mol 
H 2 mole
2 mol NH3
 5.0 mol H 2 
 3.3 mol NH3
3 mol H 2
6-6
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.
6-7
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.
b.
What is the limiting reactant?
What is the maximum amount of AlCl3, in grams,
that can be produced?
6-8
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?
2Al(s) +
3Cl2(g) 
2AlCl3(s)
53.96 g
212.7 g
266.7 g
If 0.40 g Al is reacted, then
Cl 2 g
Needed Cl 2 g  reacted Al g 
Al g
212.7 Cl 2
 0.40 g Al 
 1.6 g Cl 2  actual Cl 2 g
53.96 g Al
Hence, Cl 2 is not enough. It is limiting reatant.
6-9
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?
The amount of product depends on the amount of
limiting reactant.
6-10
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?
2Al(s) +
3Cl2(g) 
2AlCl3(s)
53.96 g
212.7 g
266.7 g
AlCl3 g
AlCl3 g  reacted Cl 2 g 
Cl 2 g
266.7 AlCl3
 0.60 g Cl 2 
 0.75 g AlCl3
212.7 g Cl 2
6-11
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
6-12
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
6-13
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?
2 mol NaCl
Theoretical yield = 0.20 mol Cl 2 ×
= 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
6-14
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
6-15
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.
6-16
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
6-17
Homework
Page 242-
Problem
6.39
6.41c
6.43
6.58
6-18