Stoichiometry

Consider : 4NH3 + 5O2 6H2O + 4NO• How many moles of H2O are produced if

0.176 mol of O2 are used?

• How many moles of NO are produced in the reaction if 17 mol of H2O are also produced?

6 mol H2O 5 mol O2

x

Stoichiometry questions (1)

# mol H2O= 0.176 mol O2 0.2112 mol H2O

=

4 mol NO 6 mol H2O

x # mol NO= 17 mol H2O11.33 mol NO

=

Notice that a correctly balanced equation is essential to get the right answer

Consider : 4NH3 + 5O2 6H2O + 4NO• How many grams of H2O are produced if 1.9

mol of NH3 are combined with excess oxygen?

• How many grams of O2 are required to produce 0.3 mol of H2O?

6 mol H2O 4 mol NH3

x

Stoichiometry questions (2)

# g H2O=1.9 mol NH3

51.4 g H2O

= 18.02 g H2O 1 mol H2O

x

5 mol O2 6 mol H2O

x

# g O2=0.3 mol H2O

8 g O2= 32 g O2 1 mol O2

x

Consider : 4NH3 + 5O2 6H2O + 4NO• How many grams of NO is produced if 12 g of

O2 is combined with excess ammonia?

4 mol NO 5 mol O2

x

Stoichiometry questions (3)

# g NO=

12 g O2

9.0 g NO=

30.01 g NO 1 mol NO

x 1 mol O2 32 g O2

x

DO NOW- Balance

• Al + HCl AlCl3 + H2

• N2 + H2 NH3

• Al + O2 Al2O3

• C3H8 + O2 CO2 + H2O

• 2Al + 6HCl 2AlCl3 + 3H2

• N2 + 3H2 2NH3

• 4Al + 3O2 2 Al2O3

• C3H8 + 5O2 3CO2 + 4H2O

Limiting Reagents:Limiting Reagents:• Do two separate calculations using both given

quantities. The smaller answer is correct.Q - How many g NO are produced if 20 g NH3 is

burned in 30 g O2? 4NH3 + 5O2 6H2O+ 4NO

4 mol NO5 mol O2

x 30 g O2

22.5 g NO=

30.0 g NO1 mol NO

x 1 mol O2 32.0 g O2

x

4 mol NO4 mol NH3

x

# g NO=20 g NH3

35.3 g NO=

30.0 g NO1 mol NO

x 1 mol NH3 17.0 g NH3

x

Limiting ReagentsLimiting Reagents

MgCl2 + 2AgNO3 Mg(NO3)2 + 2AgClIf 25 g magnesium chloride was added to 68 g silver nitrate, what mass of AgCl will be produced?

2 mol AgCl1 mol MgCl2

x

# g AgCl=25 g MgCl2

75.25 g AgCl=

143.3 g AgCl1 mol AgCl

x 1 mol MgCl295.21 g MgCl2

x

2 mol AgCl2 mol AgNO3

x

# g AgCl=68 g AgNO3

57.36 g AgCl=

143.3 g AgCl1 mol AgCl

x 1 mol AgNO3

169.88 g AgNO3

x

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Practice questions-LRPractice questions-LR1. 2Al + 6HCl 2AlCl3 + 3H2

If 25 g of aluminum was added to 90 g of HCl, what mass of H2 will be produced?

2. N2 + 3H2 2NH3: If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent?

3. What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2?

4. When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced?

Question 1Question 1

2Al + 6HCl 2AlCl3 + 3H2

If 25 g aluminum was added to 90 g HCl, what mass

of H2 will be produced?

3 mol H2

2 mol Al x # g H2= 25 g Al = 2.78 g H2

2.0 g H2

1 mol H2

x 1 mol Al27.0 g Al

x

3 mol H2

6 mol HClx # g H2 = 90 g HCl = 2.47 g H2

2.0 g H2

1 mol H2

x 1 mol HCl36.5 g HCl

x

Question 2Question 2

N2 + 3H2 2NH3

If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent?

2 mol NH3

1 mol N2 x

# g NH3=

20 g N2 = 24.3 g NH3 17.0 g NH3

1 mol NH3

x 1 mol N2

28.0 g N2

x

2 mol NH3

3 mol H2

x

# g NH3 =5.0 g H2 = 28.3 g NH3 17.0 g NH3

1 mol NH3

x 1 mol H2

2.0 g H2

x

N2 is the limiting reagent

Question 3Question 3

4Al + 3O2 2 Al2O3

What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2?

2 mol Al2O3

4 mol Al x

# g Al2O3=

10.0 g Al = 18.9 g Al2O3 102.0 g Al2O3

1 mol Al2O3

x 1 mol Al27.0 g Al

x

2 mol Al2O3

3 mol O2

x

# g Al2O3=

20.0 g O2 = 42.5 g Al2O3 102.0 g Al2O3

1 mol Al2O3

x 1 mol O2

32.0 g O2

x

Question 4Question 4C3H8 + 5O2 3CO2 + 4H2O

When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g

of O2, how much CO2 is produced?3 mol CO2

1 mol C3H8 x

# g CO2=15.0 g C3H8 = 45.0 g CO2

44.0 g CO2

1 mol CO2

x 1 mol C3H8

44.0 g C3H8

x

3 mol CO2

5 mol O2

x

# g CO2=60.0 g O2 = 49.5 g CO2

44.0 g CO2

1 mol CO2

x 1 mol O2

32.0 g O2

x

5. Limiting reagent questions give values for two or more reagents (not just one)

Percent YieldQ - 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

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 QuestionsThe 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?

1. 107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield.

2KClO3 2KCI + 3O2

2. 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 H2O

x

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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