Limiting Reactants
(Reagents)
and Percent Yield
Calculations need to be based on
the limiting reactant.
• Example 1: Suppose a box contains 87
bolts, 110 washers and 99 nails. How
many sets of 1 bolt, 2 washers and 1
nail can you use to create? What is
the limiting factor?
55 sets; washers limit the amount
Calculations need to be based on
the limiting reactant.
• Example 2: What is the maximum
mass of sulfur dioxide that can be
produced by the reaction of 95.6 g
carbon disulfide with 100. g oxygen?
CS2 + O2  CO2 + SO2
Start by balancing the equation…
Calculations need to be based on
the limiting reactant.
• Example 2: What is the maximum
mass of sulfur dioxide that can be
produced by the reaction of 95.6 g
carbon disulfide with 100. g oxygen?
CS2 + 3O2  CO2 + 2SO2
Now solve the problem…
Calculations need to be based on
the limiting reactant.
• Example 2: What is the maximum mass of sulfur dioxide that
can be produced by the reaction of 95.6 g carbon disulfide
with 100. g oxygen?
CS2 + 3O2  CO2 + 2SO2
95.6 g CS 2
1m ol CS 2 2m ol SO2 64.1g SO2



 161g SO2
76.2 g CS 2 1m ol CS 2 1m ol SO2
100.g O2
1m ol O2 2m ol SO2 64.1g SO2



 134g SO2
32.0 g O2 3m ol O2 1m ol SO2
Which reactant is LIMITING?
Calculations need to be based on
the limiting reactant.
• Example 2: What is the mass of sulfur dioxide that can be
produced by the reaction of 95.6 g carbon disulfide with
100. g oxygen?
CS2 + 3O2  CO2 + 2SO2
95.6 g CS 2
1m ol CS 2 2m ol SO2 64.1g SO2



 161g SO2
76.2 g CS 2 1m ol CS 2 1m ol SO2
100.g O2
1m ol O2 2m ol SO2 64.1g SO2



 134g SO2
32.0 g O2 3m ol O2 1m ol SO2
O2 limits the amount of SO2 that can be
produced. CS2 is in excess.
Calculations need to be based on
the limiting reactant.
• Example 2: What is the mass of sulfur dioxide that can be
produced by the reaction of 95.6 g carbon disulfide with
100. g oxygen?
CS2 + 3O2  CO2 + 2SO2
95.6 g CS 2
1m ol CS 2 2m ol SO2 64.1g SO2



 161g SO2
76.2 g CS 2 1m ol CS 2 1m ol SO2
100.g O2
1m ol O2 2m ol SO2 64.1g SO2



 134g SO2
32.0 g O2 3m ol O2 1m ol SO2
134 g of SO2 can be produced in this reaction.
Calculations need to be based on
the limiting reactant.
• Example 3: What mass of CO2 could
be formed by the reaction of 8.0 g
CH4 with 48 g O2?
CH4 + O2  CO2 + H2O
Start by balancing the equation…
Calculations need to be based on
the limiting reactant.
• Example 3: What mass of CO2 could
be formed by the reaction of 8.0 g
CH4 with 48 g O2?
CH4 + 2O2  CO2 + 2H2O
Now solve the problem…
Calculations need to be based on
the limiting reactant.
• Example 3: What mass of CO2 could be formed by the
reaction of 8.0 g CH4 with 48 g O2?
CH4 + 2O2  CO2 + 2H2O
8.0 g CH 4
48g O2
1mol CH 4 1mol CO2 44.0 g CO2



 22g CO2
16.0 g CH 4 1mol CH 4 1mol CO2
1mol O2 1mol CO2 44.0 g CO2



 33g CO2
32.0 g O2 2mol O2
1mol CO2
Which reactant is LIMITING?
Calculations need to be based on
the limiting reactant.
• Example 3: What mass of CO2 could be formed by the
reaction of 8.0 g CH4 with 48 g O2?
CH4 + 2O2  CO2 + 2H2O
8.0 g CH 4
48g O2
1mol CH 4 1mol CO2 44.0 g CO2



 22g CO2
16.0 g CH 4 1mol CH 4 1mol CO2
1mol O2 1mol CO2 44.0 g CO2



 33g CO2
32.0 g O2 2mol O2
1mol CO2
CH4 limits the amount of CO2 that can be
produced. O2 is in excess.
Many chemical reactions do not go
to completion (reactants are not
completely converted to products).
Percent Yield: indicates what
percentage of a desired product is
obtained.
experimental yield
% Yield 
 100
theoretical yield
• So far, the masses we have calculated
from chemical equations were based
on the assumption that each reaction
occurred 100%.
• The THEORETICAL YIELD is the
yield calculated from the balance
equation.
• The ACTUAL YIELD is the amount
“actually” obtained in an experiment.
• Look back at Example 2. We found
that 134 g of SO2 could be formed
from the reactants.
• In an experiment, you formed 130 g
of SO2. What is your percent yield?
130g
% Yield 
100  97%
134g
Example: A 10.0 g sample of ethanol, C2H5OH, was
boiled with excess acetic acid, CH3COOH, to produce
14.8 g of ethyl acetate, CH3COOC2H5. What percent
yield of ethyl acetate is this?
CH3COOH + C2H5OH  CH3COOC2H5 + H2O
10.0g C H OH 1 mol C H OH 1 mol CH COOC H
88 g CH COOC H
2 5
2 5
3
2 5
3
2 5


46g C H OH
1 mol C H OH
1 mol CH COOC H
2 5
2 5
3
2 5
 19.1g CH3COOC2H5
Example: A 10.0 g sample of ethanol, C2H5OH, was
boiled with excess acetic acid, CH3COOH, to produce
14.8 g of ethyl acetate, CH3COOC2H5. What percent
yield of ethyl acetate is this?
experimental yield
% Yield 
 100
theoretical yield
14.8 g
% Yield 
 100
19.1 g
% Yield  77.5%