Stoichiometry: Problem Sheet 1

Chemistry: Stoichiometry ? Problem Sheet 1

Name: ________________________ Hour: ____ Date: ___________

Directions: Solve each of the following problems. Show your work, including proper units, to earn full credit.

1. Silver and nitric acid react according to the following balanced equation: 3 Ag(s) + 4 HNO3(aq) 3 AgNO3(aq) + 2 H2O(l) + NO(g)

A. How many moles of silver are needed to react with 40 moles of nitric acid?

B. From the amount of nitric acid given in Part A, how many moles of silver nitrate will be produced?

C. From the amount of nitric acid given in Part A, how many moles of water will be produced?

D. From the amount of nitric acid given in Part A, how many moles of nitrogen monoxide will be made?

2. Given the balanced equation:

2 N2H4(l) + N2O4(l) 3 N2(g) + 4 H2O(g)

A. How many moles of dinitrogen tetrahydride are required to produce 57 moles of nitrogen?

B. How many moles of dinitrogen tetroxide are required to produce 57 moles of nitrogen?

C. How many moles of water are produced when 57 moles of nitrogen are made?

3. Calculate the mass of aluminum oxide produced when 3.75 moles of aluminum burn in oxygen.

Answers:

1A. 30 mol Ag 1B. 30 mol AgNO3

1C. 20 mol H2O 1D. 10 mol NO

2A. 38 mol N2H4 2B. 19 mol N2O4

2C. 76 mol H2O 3. 191 g Al2O3

4. At a very high temperature, manganese is isolated from its ore, manganomanganic oxide, via the following balanced equation: 3 Mn3O4(s) + 8 Al(s) 4 Al2O3(s) + 9 Mn(s) A. How many manganese atoms are liberated if 54.8 moles of Mn3O4 react with excess aluminum.

B. How many moles of aluminum oxide are made if 3580 g of manganomanganic oxide are consumed?

C. How many moles of manganomanganic oxide will react with 5.33 x 1025 atoms of aluminum?

D. If 4.37 moles of aluminum are consumed, how many molecules of aluminum oxide are produced?

5. Camels store the fat tristearin (C57H110O6) in the hump. Besides being a source of energy, the fat is a source of water for the camel because when the fat is burned, the following reaction occurs: 2 C57H110O6(s) + 163 O2(g) 114 CO2(g) + 110 H2O(l) A. At STP, what volume of oxygen is required to consume 0.64 moles of tristearin?

B. At STP, what volume of carbon dioxide is produced in Part A?

C. If 22.4 L of oxygen is consumed at STP, how many moles of water are produced?

D. Find the mass of tristearin required to produce 55.56 moles of water (about 1 liter of liquid water).

Answers:

4A. 9.9 x 1025 atoms Mn 4B. 20.9 mol Al2O3

4C. 33.2 mol Mn3O4 4D. 1.3 x 1024 m'cules Al2O3

5A. 1168 L O2 5B. 817 L CO2

5C. 0.675 mol H2O 5D. 899 g C57H110O6

KEY

Chemistry: Stoichiometry ? Problem Sheet 1

Directions: Solve each of the following problems. Show your work, including proper units, to earn full credit.

1. Silver and nitric acid react according to the following balanced equation: 3 Ag(s) + 4 HNO3(aq) 3 AgNO3(aq) + 2 H2O(l) + NO(g)

A. How many moles of silver are needed to react with 40 moles of nitric acid?

x mol Ag

40 mol HNO 3

3 mol Ag 4 mol HNO 3

30 mol Ag

B. From the amount of nitric acid given in Part A, how many moles of silver nitrate will be produced?

x mol AgNO3

40 mol HNO 3

3 mol AgNO3 4 mol HNO 3

30 mol AgNO3

C. From the amount of nitric acid given in Part A, how many moles of water will be produced?

x mol H2O

40 mol HNO 3

2 mol H2O 4 mol HNO 3

20 mol H2O

D. From the amount of nitric acid given in Part A, how many moles of nitrogen monoxide will be made?

x mol NO

40 mol HNO 3

1mol NO 4 mol HNO 3

10 mol NO

2. Given the balanced equation:

2 N2H4(l) + N2O4(l) 3 N2(g) + 4 H2O(g)

A. How many moles of dinitrogen tetrahydride are required to produce 57 moles of nitrogen?

x mol N2H4

57 mol N2

2 mol N2H4 3 mol N2

38 mol N2H4

B. How many moles of dinitrogen tetroxide are required to produce 57 moles of nitrogen?

x mol N2O4

57 mol N2

2 mol N2O4 3 mol N2

19 mol N2O4

C. How many moles of water are produced when 57 moles of nitrogen are made?

x mol H2O

57 mol N2

4 mol H2O 3 mol N2

76 mol H2O

3. Calculate the mass of aluminum oxide produced when 3.75 moles of aluminum burn in oxygen.

4 Al + 3 O2 2 Al2O3

3.75 mol excess

x g

x g Al2O3

3.75 mol Al 2 mol Al2O3 102 g Al2O3 4 mol Al 1mol Al2O3

191 g Al2O3

Answers:

1A. 30 mol Ag 1B. 30 mol AgNO3

1C. 20 mol H2O 1D. 10 mol NO

2A. 38 mol N2H4 2B. 19 mol N2O4

2C. 76 mol H2O 3. 191 g Al2O3

4. At a very high temperature, manganese is isolated from its ore, manganomanganic oxide, via the following balanced equation:

3 Mn3O4(s) + 8 Al(s) 4 Al2O3(s) + 9 Mn(s)

A. How many manganese atoms are liberated if 54.8 moles of Mn3O4 react with excess aluminum.

x atoms Mn

54.8 mol Mn3O4

9 mol Mn 3 mol Mn3O4

6.02 1023 atoms Mn 1 mol Mn

9.9 1025 atoms Mn

B. How many moles of aluminum oxide are made if 3580 g of manganomanganic oxide are consumed?

x mol Al2O3

3580 g Mn3O4

1mol Mn3O4 229 g Mn3O4

4 mol Al2O3 3 mol Mn3O4

20.9 mol Al2O3

C. How many moles of manganomanganic oxide will react with 5.33 x 1025 atoms of aluminum?

x mol Mn3O4

5.33

1025 atoms Al 6.02

1mol Al 1023 atoms Al

3 mol Mn3O4 8 mol Al

33.2 mol Mn3O4

D. If 4.37 moles of aluminum are consumed, how many molecules of aluminum oxide are produced?

x m' cules Al2O3

4.37 mol Al 4 mol Al2O3 8 mol Al

6.02 1023 m' cules Al2O3 1 mol Al2O3

1.3 104 m' cules Al2O3

5. Camels store the fat tristearin (C57H110O6) in the hump. Besides being a source of energy, the fat is a source of water for the camel because when the fat is burned, the following reaction occurs:

2 C57H110O6(s) + 163 O2(g) 114 CO2(g) + 110 H2O(l) A. At STP, what volume of oxygen is required to consume 0.64 moles of tristearin?

x L O2

0.64 mol C57H110 O6

163 mol O2 2 mol C57H110 O6

22.4 L CO 2 1 mol O2

B. At STP, what volume of carbon dioxide is produced in Part A?

1168 L O 2

x L CO2

0.64 mol C57H110O6

114 mol CO2 2 mol C57H110 O6

22.4 L CO 2 1 mol CO2

817 L CO 2

C. If 22.4 L of oxygen is consumed at STP, how many moles of water are produced?

x mol H2O

22.4 L O2

1mol O2 22.4 L O2

110 mol H2O 163 mol O2

0.675 mol H2O

D. Find the mass of tristearin required to produce 55.56 moles of water (about 1 liter of liquid water).

x g C57H110O6

55.6 mol H2O

2 mol C57H110O6 110 mol H2O

890 g C57H110O6 1 mol C57H110O6

899 g C57H110O6

Answers:

4A. 9.9 x 1025 atoms Mn 4B. 20.9 mol Al2O3

4C. 33.2 mol Mn3O4 4D. 1.3 x 1024 m'cules Al2O3

5A. 1168 L O2 5B. 817 L CO2

5C. 0.675 mol H2O 5D. 899 g C57H110O6

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