Stoichiometry: Problem Sheet 1
[Pages:4]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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