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CHE1031
CHE1031 Chapter 3 Practice Problem Key
These examples can be used with tutors or during help sessions.
Moles
1. If Avogadro’s number of pennies is divided equally among the 300 million citizens of the US,
how many dollars would each person receive? How does this compare with the US GDP:
$14.4 trillion in 2008?
6.02x1023 pennies
$1.00
= $2.01x1013
300x106 person
100 pennies US person
$14.4x1012 GDP = $48ooo/US person
300x106 citizens
2. Calculate these quantities:
a. mass (g) of 0.105 moles of sucrose (C12H22O11)
b. moles of Zn(NO3)2 in 143.50 g
c. number of molecules in 8.447x10-2 moles of C6H6
d. number of O atoms in 6.25x10-3 mol Al(OH)3
a. 0.105 moles 342 g = 35.9 g
1 mole
b. 143.50 g 1 mole = 0.758 mol
189.35 g
-2
c. 8.447x10 mol 6.02x1023 molecules = 5.085x1022 molecules
1 mol
-3
d. 6.25x10 mol Al(OH)3 6.02x1023 molecules 3 O atoms = 1.13x1022 O atoms
1 mol
1 molecule
3. What is the mass (g) of 1.223 mol of iron (III) sulfate?
Fe2(SO4)3 MW = 399.76 g/mol
1.223 mol
399.76 g = 488.9 g
1 mol
Molecular weight (MW) & percent composition
4. Determine the formula weights (MW) of these compounds:
a. dinitrogen monoxide (aka laughing gas)
b. benzoic acid, C7H6O2, a food preservative
c. Mg(OH)2, Phillip’s Milk of Magnesia
a. N2O = 44.01 g/mol
b. 122.11 g/mol
c. 58.30 g/mol
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CHE1031
5. Determine the formula weights (MW) of these compounds:
a. urea, (NH2)2CO, a nitrogen fertilizer
b. isopentyl acetate, CH3CO2C5H11, the odor of bananas
a. 60.06 g/mol
b. 130.19 g/mol
Empirical & molecular formulas
6. Determine the empirical formula for each of these compounds if a sample contains:
a. 0.104 mol K, 0.052 mol C, 0.156 mol O
b. 5.28 g Sn and 3.37 g F
c. 87.5% N and 12.5% H by mass
a. K = 0.104/0.052 = 2
C = 0.052/0.052 = 1
K2CO3
O = 0.156/ 0.052 = 3
b. Sn = 5.28 g 1 mol = 0.0445 mol / 0.0445 = 1
118.71 g
SnF4
F = 3.37 g 1 mol = 0.177 mol / 0.0445 = 4
19.00 g
c. N = 87.5 g 1 mol = 6.24 mol / 6.24 = 1
14.01 g
NH2
H = 12.5 g 1 mol = 12.4 mol / 6.24 = 2
1.01 g
7. Determine the empirical formulas of these compounds:
a. 55.3% K, 14.6% P and 30.1% O
b. 24.5% Na, 14.9% Si, and 60.6% F
c. 62.1% C, 5.21% H, 12.1% N, and the rest O
a. 55.3 g 1 mol = 1.41 mol K / 0.47 = 3
39.10 g
14.6 g 1 mol = 0.47 mol P / 0.47 = 1
K3PO4
30.97 g
30.1 g 1 mol = 1.88 mol O / 0.47 = 4
15.99 g
b. 24.5 g 1 mol = 1.06 mol Na / 0.53 = 2
22.99 g
14.9 g 1 mol = 0.53 mol Si / 0.53 = 1
Na2SiF6
28.08 g
60.06 g 1 mol = 3.16 mol F / 0.53 = 6
19.00 g
c. 62.1 g 1 mol = 5.17 mol C / 0.864 = 6 x 2 = 12
12.01 g
5.21 g 1 mol = 5.07 mol H / 0.864 = 6 x 2 = 12
1.01 g
C12H12N2O3
12.1 g 1 mol = 0.864 mol N / 0.864 = 1 x 2 = 2
14.01 g
20.6 g 1 mol = 1.29 mol O / 0.864 = 1.5 x 2 = 3
15.99 g
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CHE1031
Chemical reactions
8. Under the appropriate experimental conditions, H2 and CO undergo a combination
reaction to produce CH3OH. How many CO molecules are required to react with eight H2
molecules?
2H2 + CO  CH3OH
Four CO molecules would be required to combine with eight H2 molecules.
Balancing chemical equations
9. What is the difference between adding a subscript 2 to the end of CO to create CO2, and
adding a coefficient in front of the formula to give 2 CO?
The subscript 2 creates a formula with one carbon and two oxygen atoms.
The coefficient 2 means two molecules of CO, each with one C and one O.
10. Is this chemical equation consistent with the law of conservation of mass?
3 Mg(OH)2(s) + 2 H3(PO4)(aq)  Mg3(PO4)2(s) + 6 H2O(l)
Yes, it’s a balanced chemical equation.
11. Balance this equation:
(NH4)(NO3)(s)  N2(g) + O2(g) + H2O(g)
(NH4)(NO3)(s)  N2(g) + 1/2O2(g) + 2H2O(g)
12. Balance this equation:
Ca3P2(s) + H2O(l)  Ca(OH)2(aq) + PH3(g)
Ca3P2(s) + 6H2O(l)  3Ca(OH)2(aq) + 2PH3(g)
13. Balance this equation:
Al(OH)3(s) + H2(SO4)(aq)  Al2(SO4)3(aq) + H2O(l)
2Al(OH)3(s) + 3H2(SO4)(aq)  Al2(SO4)3(aq) + 6H2O(l)
14. Balance this equation:
C2H3(NH2)(g) + O2(g)  CO2(g) + H2O(g) + N2(g)
2C2H3(NH2)(g) + 13/2O2(g)  4CO2(g) + 5H2O(g) + N2(g)
15. Write a balanced chemical equation representing this: solid calcium carbide, CaC2, reacts
with water to form an aqueous solution of calcium hydroxide and acetylene gas.
CaC2 + 2H2O  Ca(OH)2 + C2H2
Patterns of chemical reactivity
16. Determine the chemical formula of the product formed when the metal aluminum
combines with the non-metal bromine, Br2. Write a balanced chemical equation for this
reaction.
2Al + 3Br2  2AlBr3
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17. What products form when a compound containing C, H and O is completely combusted in
air? Write a balanced chemical equation for the combustion of acetone, C3H6O in oxygen
gas.
C3H6O + 4O2  3CO2 + 3H2O
18. Write a balanced chemical equation for the reaction that occurs when calcium metal
undergoes a combination reaction with O2 gas.
2Ca + O2  2CaO
Stoichiometry & conversions
19. The reaction between potassium superoxide and carbon dioxide is used in self-contained
breathing equipment used by first responders and occurs as follows:
4 KO2 + 2 CO2  2 K2CO3 + 3 O2
a. How many moles of O2 are produced when 0.400 mole of KO2 reacts?
b. How many grams of KO2 are needed to form 7.50 g of O2?
c. How many grams of CO2 are used when 7.50 g of O2 are produced?
a. 0.400 mol KO2 3 mol O2 = 0.300 mol O2
4 mol KO2
b. 7.50 g O2 1 mol O2 4 mol KO2 71.08 g = 22.22 g KO2
31.98 g 3 mol O2 1 mol KO2
c. 7.50 g O2 1 mol O2 2 mol CO2 43.99 g = 6.88 g CO2
31.98 g 3 mol O2 1 mol CO2
Limiting reactants & theoretical yield
20. Theoretical and actual yield:
a. Why is actual yield almost always lower than theoretical yield?
b. Can a reaction ever have a 110% yield?
a. Actual yield is lower because life, technology, science, chemistry aren’t perfect.
b. Well, not unless some sort of error has occurred.
21. Aluminum hydroxide, Al(OH)3 reacts with sulfuric acid as follows:
2 Al(OH)3 (s) + 3 H2(SO4) (aq)  Al2(SO4)3 (aq) + 6 H2O (l)
a. Which reactant is limiting when 0.500 mol of each reacts?
b. How many moles of Al2(SO4)3 will be produced?
c. How many moles of excess reactant remain?
a. 0.500 mol 1 mol Al2(SO4)3 = 0.250 mol Al2(SO4)3
2 mol Al(OH)3
0.500 mol 1 mol Al2(SO4)3 = 0.167 mol Al2(SO4)3
limiting
3 mol H2(SO4)
a. 0.167 moles
b. 0.167 mol Al2(SO4)3 2 mol Al(OH)3 = 0.333 mol -> 0.500 – 0.333 = 0.167 mol excess
1 mol Al2(SO4)3
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