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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%?

theoretical yield

x 100%

Answers1)

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:

- 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

actual

2 mol H2O

138 g H2O

x

=

x

2.02 g H2

theoretical

143 g H2O

2 mol H2

18.02 g H2O

x

1 mol H2O

=

=

143 g

96.7%

Sample problemQ - 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:

# g H2O=

16 g H2

Step 3: Calculate % yield

% yield =

x 100%

x 100%

2 mol NH3

40.5 g NH3

x

=

theoretical

3 mol H2

227 g NH3

17.04 g NH3

x

1 mol NH3

=

=

17.8%

227 g

Practice problemQ - 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:

# g NH3=

20.0 mol H2

Step 3: Calculate % yield

% yield =

x 100%

x 100%

actual

1 mol KClO3

3 mol O2

78.5% x 1.958 g O2

x g O2

1.78 g O2

x

=

x

=

=

122.55 g KClO3

theoretical

theoretical

1.958 g O2

1.958 g O2

100%

2 mol KClO3

32 g O2

x

1 mol O2

=

=

=

=

78.5%

90.9%

1.537 g O2

1.958 g

Answers# g O2= (also works if you use mol O2)

5.00 g KClO3

4) 2KClO3 2KCl + 3O2

a)

b)

c)

% yield =

x 100%

x 100%

% yield =

x 100%

x 100%

x g O2

actual

1 mol H2

2 mol H2O

2 mol H2O

58 g H2O

x

x

x

x

=

32 g O2

2.02 g H2

theoretical

1 mol O2

2 mol H2

62.4 g H2O

18.02 g H2O

18.02 g H2O

x

x

1 mol H2O

1 mol H2O

=

=

92.9%

68 g

=

62.4 g

Challenging question2H2 + 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

# g H2O=

60 g O2

# g H2O=

7.0 g H2

% yield =

x 100%

x 100%

More Percent Yield Questions

Note: try “shortcut” for limiting reagent problems

- The electrolysis of water forms H2 and O2. 2H2O 2H2 + O2 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

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

actual

1 mol O2

12.3 g O2

x

=

x

18.02 g H2O

theoretical

12.43 g O2

2 mol H2O

32 g O2

x

1 mol O2

=

=

12.43 g

98.9%

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=

14.0 g H2O

Finally, calculate % yield

% yield =

x 100%

x 100%

actual

3 mol O2

107 g O2

x

=

x

122.55 g KClO3

theoretical

117.5 g O2

2 mol KClO3

32 g O2

x

1 mol O2

=

=

117.5 g

91.1%

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=

300 g KClO3

Finally, calculate % yield

% yield =

x 100%

x 100%

1 mol FeS

220 g O2

x

=

theoretical

1 mol Fe

263.7 g O2

87.91 g FeS

x

1 mol FeS

=

=

83.4%

263.7 g

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

# g FeS=

3.00 mol Fe

Finally, calculate % yield

% yield =

x 100%

x 100%

143 g Fe2O3

=

theoretical

181.48 g Fe2O3

=

=

=

181.48 g Fe2O3

78.8%

199.7 g Fe2O3

1 mol O2

1 mol FeS2

2 mol Fe2O3

2 mol Fe2O3

159.7 g Fe2O3

159.7 g Fe2O3

x

x

x

x

x

x

32 g O2

119.97 g FeS2

4 mol FeS2

11 mol O2

1 mol Fe2O3

1 mol Fe2O3

- 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=

300 g FeS2

200 g O2

% yield =

x 100%

x 100%

42% x 57.08 g Cl2

x g Cl2

=

=

theoretical

100%

57.08 g Cl2

=

=

=

=

42%

57.08 g Cl2

24 g Cl2

62.13 g Cl2

1 mol MnO2

1 mol Cl2

1 mol Cl2

71 g Cl2

70.9 g Cl2

x

x

x

x

x

86.94 g MnO2

4 mol HCl

1 mol MnO2

1 mol Cl2

1 mol Cl2

# g Cl2=

- 70 g of MnO2 + 3.5 mol HCl gives a 42% yield. How many g of Cl2 is produced? MnO2 + 4HCI MnCl2 + 2H2O + Cl2

70 g MnO2

3.5 mol HCl

% yield =

x 100%

x 100%

x g Cl2

For more lessons, visit www.chalkbored.com

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