CHM134 General Chemistry I
Exam 1, Fall 2009 – Drs. Foy and Steel
Name
SOLUTIONS
Section 1: Fill in the blanks as instructed in each question.
1. (8 points)Balance each equation by providing the appropriate coefficients. If you leave a
blank before a compound it will be interpreted as a coefficient of 1.
2 C6H6 + 15 O2 → 12 CO2 +
6 H2O
3 Br2 + 8 NH3 → 6 NH4Br +
2 CH3OH +
2
CON2H4 →
3 O2 →
1
CN2H2 +
2 CO2 +
2
N2
4 H2O
NH3 +
1
CO2
2. (4 points) Indicate whether each species is soluble or insoluble in water.
soluble
KOH
soluble
Pb(NO3)2
insoluble
insoluble
Cu(OH)2
insoluble
soluble
CuCl2
FeO
insoluble
FePO4
PbI2
soluble
KI
3. (4 points) Indicate the oxidation number of each element in these species.
PO43-
CH2Cl2
0 C
+1 H
-1 Cl
+5 P -2 O
NaBrO4
+1 Na
+7 Br -2 O
4. (3 points) Indicate the gas that is most likely to be produced from each reaction.
CO2
NaHCO3(aq) + CH3CO2H(aq) → ????
H2S
K2S(aq) + 2 HCl(aq) → ????
NH3
NH4Br(aq) + KOH(aq) → ????
5. (6 points) Indicate whether each reaction is a precipitation, acid-base, gas forming, or
redox reaction.
Precipitation
2 NaI(aq) + Pb(NO3)2(aq) → 2 NaNO3(aq) +
PbI2(s)
Gas Forming
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)
Redox
Acid-Base
Precipitation
LiOH(aq) + HCl(aq) → H2O(l) + LiCl(aq)
HCl(aq) + AgNO3(aq) → HNO3(aq) + AgCl(s)
2 H2O(l) → 2 H2(g) + O2(g)
Redox
6. (5 points) Balance this redox reaction and indicate which species is the oxidizing agent
and which is the reducing agent.
K2Cr2O7(aq) +
K2Cr2O7
14 HI(aq) → 2 KI(aq) + 2 CrI3(aq) +
oxidizing agent
HI
3 I2(s) + 7 H2O(l)
reducing agent
Section 2: Complete each question. You must show your work in order to earn full credit.
7. (10 points) Nitric oxide is made from the oxidation of ammonia. How many moles of
nitric oxide can be made from the reaction of 3.80 mol NH3 with 4.25 mol O2? How
many moles of the excess reactant will be leftover?
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
3.80 mol NH 3 ×
4 mol NO
= 3.80 mol NO
4 mol NH 3
4 mol NO
= 3.40 mol NO
5 mol O 2
Therefore, only 3.40 moles of NO will be produced
The limiting reactant is the O2, thus
4 mol NH 3
4.25 mol O 2 ×
= 3.40 mol NH 3 are used, and
5 mol O 2
3.80 – 3.40 = 0.40 mol NH3 will be leftover
4.25 mol O 2 ×
8. (10 points) What mass of carbon reacts completely with 17.8 grams of SiO2 according to
the following balanced equation?
SiO2(s) + 3 C(s) → SiC(s) + 2 CO(g)
17.8 g SiO 2 ×
1 mol SiO 2
3 mol C
12.01 g C
×
×
= 10.7 g C
60.09 g SiO 2 1 mol SiO 2 1 mol C
9. (10 points) The density of a 20.0% by mass ethylene glycol (C2H6O2) solution in water is
1.03 g/mL. Find the molarity of the solution.
You must assume a volume, say 100 mL, of the solution:
100
1.03
0.20 103
20.6
1
62.07
103
20.6
0.3319
0.3319
0.100
3.32
10. (10 points) If 5.00 mL of 0.314-M KOH is diluted to exactly 125 mL with water, what is
the concentration of the resulting solution?
M 1 V1 = M 2 V2
M2 =
M 1 V1
V2
(0.314 M )(0.00500 L)
(0.125 L)
= 0.0126 M
=
11. (10 points) The balanced equation for the combustion of octane appears below. If your
car converts octane to CO2 and water with a typical percent yield of 82.4%, what mass of CO2
does your car emit for every gallon of octane your burn?
1 gallon = 3.7854 L, the density of octane is 0.703 g/mL
2 C8H18(g) + 25 O2(g) → 18 H2O(s) + 16 CO2(g)
Find the mass of octane in a gallon:
0.703
3785.4
2661
Now find the mass of CO2 produced:
1
114.22
2661
16
2
44.01
1
8200
8.20
Since the reaction has an 82.4% yield:
0.824 8.20
6.76
11. (10 points) Combustion analysis of a 13.42-g sample of equilin (which contains only
carbon, hydrogen, and oxygen) produced 39.61 g of CO2 and 9.01 g of H2O. Find the
empirical formula of equilin.
1
44.01
39.61
9.01
12.01
1
0.900
1
18.02
2
1
0.5
10.81
1.008
1
1.008
13.42 g sample – 10.81 g C – 1.008 g H = 1.60 g O
1.60
0.900
0.100
9
1
16.00
1.00
0.100
0.100
10
0.100
0.100
The empirical formula is C9H20O
1
13. (10 points) Assuming the reaction below occurs with an 85.8% yield, what volume of
2.00 M H3PO4 should you use to ensure that you produce 25.0 g of hydrogen gas?
2 H3PO4(aq) + 3 Mg(s) → 3 H2(g) + Mg3(PO4)2(s)
Find the theoretical yield:
25.0
0.858
25.0
0.858
29.1
Now find the volume of solution required to produce this quantity of H2:
29.1
1
2.016
2
3
1
2.00
4.82
Bonus: Circle and identify any functional groups (other than alkanes) in the structures
below.
alkene
O
O
OH
ketone
carboxylic acid
OH
alkyne
NH2
amine
O
ether
alcohol