1. Percentage Yield

2. Assignment: Read 5.6 up to sample (238-9)
Define the following terms: yield, theoretical yield, actual yield, percentage yield.
Based on your reading, give 4 reasons why the actual yield in a chemical reaction often falls short of the theoretical yield.
Read the sample problem on the next slide and try the practice problem on slide number 5
When 5.00 g of KClO3 is heated it decomposes according to the equation: 2KClO3  2KCl + 3O2
a) Calculate the theoretical yield of oxygen.
b) Give the % yield if 1.78 g of O2 is produced.
c) How much O2 would be produced if the percentage yield was 78.5%?

3. Answers 1) Yield: the amount of product
Theoretical yield: the amount of product we expect, based on stoichiometric calculations
Actual yield: amount of product from a procedure or experiment (this is given in the question)
Percent yield:
x 100%
actual yield
theoretical yield 2)
Not all product is recovered (e.g. spattering)
Reactant impurities (e.g. weigh out 100 g of chemical which has 20 g of junk)
A side reaction occurs (e.g. MgO vs. Mg3N2)
The reaction does not go to completion

4. Sample problem
Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2?
Step 1: write the balanced chemical equation
2H2 + O2  2H2O
Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:
1 mol H2
2.02 g H2 x
2 mol H2O
2 mol H2 x
18.02 g H2O
1 mol H2O x
# g H2O=
16 g H2
143 g =
Step 3: Calculate % yield
actual
theoretical
138 g H2O
143 g H2O =
% yield =
x 100%
x 100%
96.7% =

5. Practice problem
Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2?
Step 1: write the balanced chemical equation
N2 + 3H2  2NH3
Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:
2 mol NH3
3 mol H2 x
17.04 g NH3
1 mol NH3 x
# g NH3=
20.0 mol H2
227 g =
Step 3: Calculate % yield
actual
theoretical
40.5 g NH3
227 g NH3 =
% yield =
x 100%
x 100%
17.8% =

6. Answers 4) 2KClO3  2KCl + 3O2 a) b) c)
# g O2= (also works if you use mol O2)
1 mol KClO3
g KClO3 x
3 mol O2
2 mol KClO3 x
32 g O2
1 mol O2 x
5.00 g KClO3
1.958 g =
actual
theoretical
1.78 g O2
1.958 g O2 =
% yield =
x 100%
x 100%
90.9% =
actual
theoretical
x g O2
1.958 g O2 =
% yield =
x 100%
x 100%
78.5% =
78.5% x g O2
100% =
x g O2
1.537 g O2 =

7. Challenging question 2H2 + O2  2H2O
What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2?
Hint: determine limiting reagent first
1 mol O2
32 g O2 x
2 mol H2O
1 mol O2 x
18.02 g H2O
1 mol H2O x
# g H2O=
60 g O2
68 g =
1 mol H2
2.02 g H2 x
2 mol H2O
2 mol H2 x
18.02 g H2O
1 mol H2O x
# g H2O=
7.0 g H2
62.4 g =
actual
theoretical
58 g H2O
62.4 g H2O =
% yield =
x 100%
x 100%
92.9% =

8. More Percent Yield Questions
Note: try “shortcut” for limiting reagent problems
The electrolysis of water forms H2 and O H2O  2H2 + O What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O?
107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield. 2KClO3  2KCI + 3O2
What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulphide? Fe + S  FeS

9. More Percent Yield Questions
Iron pyrites (FeS2) reacts with oxygen according to the following equation:
4FeS2 + 11O2  2Fe2O3 + 8SO2
If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide?
70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl2 will be produced from this reaction if the % yield for the process is 42%?
MnO2 + 4HCI  MnCl2 + 2H2O + Cl2

10. Q1 The electrolysis of water forms H2 & O2. 2H2O  2H2 + O2
Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O?
Actual yield is given: 12.3 g O2
Next, calculate theoretical yield
# g O2=
1 mol H2O
18.02 g H2O x
1 mol O2
2 mol H2O x
32 g O2
1 mol O2 x
14.0 g H2O
12.43 g =
Finally, calculate % yield
actual
theoretical
12.3 g O2
12.43 g O2 =
% yield =
x 100%
x 100%
98.9% =

11. Q2
107 g of oxygen is produced by heating 300 grams of potassium chlorate.
2KClO3  2KCI + 3O2
Actual yield is given: 107 g O2
Next, calculate theoretical yield
# g O2=
1 mol KClO3
g KClO3 x
3 mol O2
2 mol KClO3 x
32 g O2
1 mol O2 x
300 g KClO3
117.5 g =
Finally, calculate % yield
actual
theoretical
107 g O2
117.5 g O2 =
% yield =
x 100%
x 100%
91.1% =

12. Q3
What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide?
Fe + S  FeS
Actual yield is given: 220 g FeS
Next, calculate theoretical yield
1 mol FeS
1 mol Fe x
87.91 g FeS
1 mol FeS x
# g FeS=
3.00 mol Fe
263.7 g =
Finally, calculate % yield
actual
theoretical
220 g O2
263.7 g O2 =
% yield =
x 100%
x 100%
83.4% =

13. First, determine limiting reagent
4FeS2 + 11O2  2Fe2O3 + 8SO2
If 300 g of FeS2 is burned in 200 g of O2, 143 g Fe2O3 results. % yield Fe2O3?
First, determine limiting reagent
# g Fe2O3=
1 mol FeS2
g FeS2 x
2 mol Fe2O3
4 mol FeS2 x
159.7 g Fe2O3
1 mol Fe2O3 x
300 g FeS2
199.7 g Fe2O3 =
1 mol O2
32 g O2 x
2 mol Fe2O3
11 mol O2 x
159.7 g Fe2O3
1 mol Fe2O3 x
200 g O2
g Fe2O3 =
actual
theoretical
143 g Fe2O3
g Fe2O3 =
% yield =
x 100%
x 100%
78.8% =

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70 g of MnO mol HCl gives a 42% yield. How many g of Cl2 is produced? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2
# g Cl2=
1 mol MnO2
86.94 g MnO2 x
1 mol Cl2
1 mol MnO2 x
70.9 g Cl2
1 mol Cl2 x
70 g MnO2
57.08 g Cl2 =
1 mol Cl2
4 mol HCl x
71 g Cl2
1 mol Cl2 x
3.5 mol HCl
62.13 g Cl2 =
actual
theoretical
x g Cl2
57.08 g Cl2 =
% yield =
x 100%
x 100%
42% =
42% x g Cl2
100% =
x g Cl2
24 g Cl2 =
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