Chemistry 1
Volume 2
Worksheet 12
Percent Yield in Chemical Reactions
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1. Calculate the percent yield of a reaction that had a theoretical yield of 3.76 g and an actual
yield of 1.45 g.
2. If 2.5 g of K3PO4 was produced in the reaction below and the percent yield was 45%, what
was the theoretical yield?
3 KOH + H3PO4 à K3PO4 + 3 H2O
3. If the actual yield of a reaction was 0.567 g, and the theoretical yield was 0.750 g, what was
the percent yield of this reaction?
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4. If the percent yield of a reaction was 65.0% and its theoretical yield was 0.652 g of CO2 what
was the actual yield?
5. Calculate the theoretical yield of a reaction that produced 4.5 g of a product with a 38%
yield.
6. Calculate the percent yield of a reaction that produced 5.78 g NH4, if its theoretical yield
was 6.78 g.
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7. Calculate the percent yield of a reaction that produced 0.350 mol HCl if the theoretical yield
was 15.36 g.
8. Calculate the percent yield of H2 of a reaction that produced 0.050 moles H2 from 4.5 g HCl
and excess Al in the reaction below:
2 Al + 6 HCl à 2 AlCl3 + 3 H2
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9. Use the following equation to answer the questions that follow.
Fe + H2SO4 à Fe2(SO4)3 + H2
a. Balance the equation.
b. Calculate the theoretical yield of Fe2(SO4)3 if 1.4 g Fe and 3.4 g H2SO4 are reacted.
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c. If the percent yield of Fe2(SO4)3 in the reaction was 67%, what was the actual yield?
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Answer Key
1. Calculate the percent yield of a reaction that had a theoretical yield of 3.76 g and an actual
yield of 1.45 g.
To answer this, we just use the following equation:
!"#$%& ()*&+
Percent yield = ,-*./*#)"%& ()*&+ x 100
Substitute what we know (theoretical yield and actual yield).
0.23 4
Percent yield = 5.67 4 x 100
Don’t forget to multiply the decimal by 100 to convert it to a percentage.
Percent yield: 0.386 x 100 = 38.6%
Correct answer 38.6%
2. If 2.5 g of K3PO4 was produced in the reaction below and the percent yield was 45%, what
was the theoretical yield?
3 KOH + H3PO4 à K3PO4 + 3 H2O
!"#$%& ()*&+
Percent yield = ,-*./*#)"%& ()*&+ x 100
We just need to substitute what we know (percent yield and actual yield) into the
equation and solve for theoretical yield. Convert percent yield to a decimal so that we
can work with it.
8.3 4
45% = ,-*./*#)"%& ()*&+ x 100
8.3 4
0.45 = ,-*./*#)"%& ()*&+
Rearrange and solve for theoretical yield.
Theoretical yield =
8.3 4
9.23
Theoretical yield = 5.6 g
Correct answer: 5.6 g
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3. If the actual yield of a reaction was 0.567 g, and the theoretical yield was 0.750 g, what was
the percent yield of this reaction?
Percent yield =
!"#$%& ()*&+
,-*./*#)"%& ()*&+
x 100
9.376 4
Percent yield = 9.639 4 x 100
Percent yield = 0.756 X 100 = 75.6 %
Correct answer: 75.6%
4. If the percent yield of a reaction was 65.0% and its theoretical yield was 0.652 g of CO2 what
was the actual yield?
!"#$%& ()*&+
Percent yield = ,-*./*#)"%& ()*&+ x 100
65% =
!"#$%& ()*&+
9.738 4
x 100
Actual yield = (0.65)(0.652 g)
Actual yield = 0.424 g
Correct answer: 0.424 g
5. Calculate the theoretical yield of a reaction that produced 4.5 g of a product with a 38%
yield.
!"#$%& ()*&+
Percent yield = ,-*./*#)"%& ()*&+ x 100
2.3 4
38% = ,-*./*#)"%& ()*&+ x 100
Theoretical yield =
2.3 4
9.5:
Theoretical yield = 12 g
Correct answer: 12 g
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6. Calculate the percent yield of a reaction that produced 5.78 g NH4, if its theoretical yield
was 6.78 g.
5.78 g NH4 is the actual yield. Any time you see “yield” or “produced,” assume that is
the actual yield.
3.6: 4
Percent yield = 7.6: 4 x 100
Percent yield = 0.853 x 100 = 85.3%
Correct answer: 85.3%
7. Calculate the percent yield of a reaction that produced 0.350 mol HCl if the theoretical yield
was 15.36 g.
To answer this, we need to convert the mol HCl into grams of HCl so that the units are
the same and cancel in the percent yield calculation.
0.350 mol HCl
36.46 g HCl
1 mol HCl
= 12.8 g HCl
08.: 4
Percent yield = 03.57 4 x 100
Percent yield = 0.833 x 100 = 83.3 %
Correct answer: 83.3%
8. Calculate the percent yield of H2 of a reaction that produced 0.050 moles H2 from 4.5 g HCl
and excess Al in the reaction below:
2 Al + 6 HCl à 2 AlCl3 + 3 H2
First, we need to calculate our theoretical yield. Since Al is in excess, HCl is the limiting
reagent. Since the actual yield of H2 is given in moles in the problem, we can save
ourselves time and just convert 4.5 g HCl into moles H2.
4.5 g HCl
1 mol HCl
36.46 g HCl
3 mol H2
6 mol HCl
9.939 ;.& <
Percent yield = 9.978 ;.& <!
!
Correct answer: 81%
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= 0.062 mol H2
9. Use the following equation to answer the questions that follow.
Fe + H2SO4 à Fe2(SO4)3 + H2
a. Balance the equation.
Fe + H2SO4 à Fe2(SO4)3 + H2
Reactants
Fe
1
H
2
SO4 1
Products
Fe
2
H
2
SO4 3
Step 1:
2 Fe + H2SO4 à Fe2(SO4)3 + H2
Reactants
Products
Fe
2
Fe
2
H
2
H
2
SO4 1
SO4 3
Step 2:
2 Fe + 3 H2SO4 à Fe2(SO4)3 + H2
Reactants
Products
Fe
2
Fe
2
H
6
H
2
SO4 3
SO4 3
Step 3:
2 Fe + 3 H2SO4 à Fe2(SO4)3 + 3 H2
Reactants
Products
Fe
2
Fe
2
H
6
H
6
SO4 3
SO4 3
Correct answer: 2 Fe + 3 H2SO4 à Fe2(SO4)3 + 3 H2
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b. Calculate the theoretical yield of Fe2(SO4)3 if 1.4 g Fe and 3.4 g H2SO4 are reacted.
1.4 g Fe
1 mol Fe
1 mol Fe2(SO4)3
399.88 g Fe2(SO4)3
55.845 g Fe
2 mol Fe
1 mol Fe2(SO4)3
3.4 g H2SO4
= 5.0 g Fe2(SO4)3
1 mol H2SO4
1 mol Fe2(SO4)3
399.88 g Fe2(SO4)3
98.079 g H2SO4
3 mol H2SO4
1 mol Fe2(SO4)3
= 4.6 g Fe2(SO4)3
Correct answer: 4.6 g Fe2(SO4)3
c. If the percent yield of Fe2(SO4)3 in the reaction was 67%, what was the actual yield?
67% =
!"#$%& =)*&+
2.78 4
x 100
(0.67)(4.6 g) = 3.1 g
Correct answer: 3.1 g
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