Assignment: Read 5.6 up to sample (238-9)1. Define the following terms: yield, theoretical

yield, actual yield, percentage yield.2. Based on your reading, give 4 reasons why the

actual yield in a chemical reaction often falls short of the theoretical yield.

3. Read the sample problem on the next slide and try the practice problem on slide number 5

4. 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%?

Answers1) Yield: the amount of productTheoretical yield: the amount of product we

expect, based on stoichiometric calculationsActual yield: amount of product from a procedure

or experiment (this is given in the question)

Percent yield:x 100%

actual yieldtheoretical 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

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 2H2OStep 2: determine actual and theoretical yield.

Actual is given, theoretical is calculated:

2 mol H2O 2 mol H2

x # g H2O= 16 g H2 143 g=

18.02 g H2O1 mol H2O

x 1 mol H2

2.02 g H2

x

Step 3: Calculate % yield138 g H2O 143 g H2O

= % yield = x 100% 96.7%= actualtheoretical x 100%

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:

2 mol NH3 3 mol H2

x # g NH3= 20.0 mol H2 227 g=

17.04 g NH3

1 mol NH3

x

Step 3: Calculate % yield40.5 g NH3 227 g NH3

= % yield = x 100% 17.8%= actualtheoretical x 100%

Answers4) 2KClO3 2KCl + 3O2a)

b)

c)

3 mol O2 2 mol KClO3

x

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

5.00 g KClO3

1.958 g=

32 g O2

1 mol O2

x 1 mol KClO3

122.55 g KClO3

x

1.78 g O2 1.958 g O2

= % yield = x 100% 90.9%= actualtheoretical x 100%

x g O2

1.958 g O2

= % yield = x 100% 78.5%= actualtheoretical x 100%

x g O278.5% x 1.958 g O2

100%= 1.537 g O2=

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

2 mol H2O 2 mol H2

x # g H2O= 7.0 g H2 62.4 g=

18.02 g H2O1 mol H2O

x 1 mol H2

2.02 g H2

x

58 g H2O 62.4 g H2O

= % yield = x 100% 92.9%= actualtheoretical x 100%

2 mol H2O 1 mol O2

x # g H2O= 60 g O2 68 g=

18.02 g H2O1 mol H2O

x 1 mol O2

32 g O2

x

More Percent Yield QuestionsNote: try “shortcut” for limiting reagent problems

1. 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?

2. 107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield.

2KClO3 2KCI + 3O2

3. 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 Questions4. 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?

5. 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

Q11. 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 yield1 mol O2

2 mol H2Ox

# g O2=14.0 g H2O 12.43 g= 32 g O2

1 mol O2

x 1 mol H2O18.02 g H2Ox

Finally, calculate % yield12.3 g O2 12.43 g O2

= % yield = x 100% 98.9%= actualtheoretical x 100%

Q22. 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

3 mol O2 2 mol KClO3

x

# g O2=

300 g KClO3

117.5 g=

32 g O2

1 mol O2

x 1 mol KClO3

122.55 g KClO3

x

Finally, calculate % yield107 g O2

117.5 g O2

= % yield = x 100% 91.1%= actualtheoretical x 100%

Q33. 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 FeS1 mol Fe

x # g FeS=3.00 mol Fe 263.7 g= 87.91 g FeS1 mol FeS

x

Finally, calculate % yield220 g O2

263.7 g O2

= % yield = x 100% 83.4%= actualtheoretical x 100%

4. 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

2 mol Fe2O3

11 mol O2 x 200 g O2

181.48 g Fe2O3=

159.7 g Fe2O3

1 mol Fe2O3

x 1 mol O2 32 g O2

x

2 mol Fe2O3

4 mol FeS2 x

# g Fe2O3=

300 g FeS2

199.7 g Fe2O3=

159.7 g Fe2O3

1 mol Fe2O3

x 1 mol FeS2 119.97 g FeS2

x

143 g Fe2O3 181.48 g Fe2O3

= % yield = x 100% 78.8%= actualtheoretical x 100%

5. 70 g of MnO2 + 3.5 mol HCl gives a 42% yield. How many g of Cl2 is produced? MnO2 + 4HCI MnCl2 + 2H2O + Cl2

1 mol Cl2

4 mol HClx 3.5 mol HCl 62.13 g Cl2= 71 g Cl2

1 mol Cl2

x

1 mol Cl2

1 mol MnO2 x

# g Cl2=

70 g MnO2

57.08 g Cl2=

70.9 g Cl2

1 mol Cl2

x 1 mol MnO2 86.94 g MnO2

x

x g Cl2

57.08 g Cl2

= % yield = x 100% 42%= actualtheoretical x 100%

x g Cl242% x 57.08 g Cl2

100%= 24 g Cl2=

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