Chemistry 100.12
Problem Session 1a
G. Marangoni
Date: September 27, 2011.
1. The density of air is 1.19 g/L. Assume that NH351 measures 150 x 16 x 50 ft?
Calculate the mass of air in NH 351.
Calculate the volume of the room.
3
 0.3048 m 
3
Vroom  150. ft 16 ft  50 ft  
  3398 m
ft


1000 L
Vroom  L   3398 m3 
 3.398 106 L
3
1m
 mass of air  3.398 106 L 1.19 g
L
6
3
 4.04 10 g  4.04 10 kg
2. A 30.7 L sample of chloroform at 20 ° C has a mass of 45.52 kg. What is the density
of chloroform at 20°C, in grams per milliliter? In pounds per US gallon?
m 45.52 kg
1L
1000 g
d 
 1.48 kg 

 1.48 g
L
mL
V
30.7 L
1000 mL 1 kg
 lb 
1lb
3785mL
 g 
d

 12.3 lb
d

US gal
 mL  453.6 g US gal
 US gal 
3. The present world consumption of oil is 84.5 million barrels. Assume a hotel pool
averages 12.0 m x 6.0 m x 2.5 m. Calculate the number of hotel pools swimming
pools that can be filled with this much crude. (1 barrel = 159.0 L).
Note – calculate the volume of an “average” hotel pool.
Vpool  12.0 m  6.0 m  2.5 m  180. m3
1000 L
 1.80 105 L
3
1m
Now calculate the volume of oil consumed per day
159.0 L
Vcrude  84.5 106 bbl 
 1.34 1010 L
bbl
1.34 1010 L
 number of pools 
 7.5 104 pools
5
1.80 10 L
Vpool  L   180. m3 
4. Complete the following table.
Substance
30
Fe3+
19 -
Protons
Neutrons
Electrons
Mass Number
Charge
14
16
14
29
0
26
30
23
56
+3
9
10
10
19
-1
Si
F
Ca 2
20
20
18
40
2+
40
20
5. Balance the following equations.
a) Ca3P2 (s) + 6 H2O (l)  3 Ca(OH)2 (aq) + 2 PH3 (g)
b) 2 AgNO3 (aq) + Na2SO4  Ag2SO4 (s) + 2 NaNO3 (aq)
c) C6H12 (g) + 9 O2 (g)  6 CO2 (g) + 6 H2O (l)
6. Obtain the empirical formula of the following substance.
58.79 % C, 9.88 % H; 31.32 % O.
Take exactly 100.00 g of Compound
nC 
58.79 g C
9.88 g H
 4.90 mol C; nH 
 9.78 mol H
12.01 g / mol
1.01 g / mol
31.32 g O
 1.96 mol O
16.00 g / mol
Next – obtain the empirical formula by dividing through by the smallest mole number
nC
nH
4.90 mol C
9.78 mol H

 2.50 C;

 5.00 H
1.96 mol
1.96 mol
1.96 mol
1.96 mol
nO
1.96 mol O

 1.00 mol O
1.96 mol
1.96 mol
2.50 C; 5.00 H ;1.00 O  2C;5H ;1O
nO 
 C2.5 H 5O1  2  C5 H10O2
7. Methanol (CH3OH) is often used as a fuel or a fuel supplement in internal combustion
engines. Methanol can be prepared industrially by the reaction of carbon monoxide
and hydrogen. Suppose we mix 426 g of CO and 14.2 moles of H2, calculate
a) The amount of methanol expected.
b) The percentage yield of methanol given that 144 g of methanol was obtained in
the reaction vessel.
a) Write down the balanced equation
CO  g   2H 2  g   CH3OH  l 
Obtain the limiting reagent
nCO 
426 g CO
 15.21 mol CO; nH 2  14.2 mol
28.01 g / mol
nH 14.2 mol
nCO 15.2 mol

15.2 mol ; 2 
 7.1 mol
1
1
2
2
 H 2 is the limiting reagent!
1CH 3OH
1
nCH3OH  nH 2 
 14.2 mol   7.10 mol
2H 2
2
mass of CH 3OH  7.10 mol 
b)
32.05 g
 227.6 g  228 g
1 mol
Calculate the percentage yield
actual yield
144 g
% yield 
100% 
100%  63.2% yield
theoretical yield
228 g
8. Cryolite, Na3AlF6, is an important industrial reagent; is made by the reaction below.
Al2O3(s) + 6 NaOH (aq) + 12 HF (g)  2 Na3AlF6 (s) + 9 H2O (l)
In an experiment, 7.81 g Al2O3 and 5.64 grams of HF (g) were dissolved in an
aqueous solution of sodium hydroxide. If 3.42 g Na3AlF6 was obtained, then what is
the percent yield for this experiment?
Obtain the limiting reagent
nAl2O3 
17.81 g Al2O3
5.64 g HF
 0.174 mol Al2O3 ; nHF 
 0.282 mol HF;
102.1 g / mol
20.01 g / mol
nAl2O3
n
0.174 mol
0.282 mol
 0.174 mol ; HF 
 0.0235 mol
1
1
12
12
 HFis the limiting reagent!
2 Na3 AlF6
1
nNa3 AlF6  nHF 
 0.282 mol   0.0470 mol
12 HF
6
209.97 g
mass of Na3 AlF6  0.0470mol 
 9.87 g
1 mol
Calculate the percentage yield
actual yield
3.42 g
% yield 
100% 
100%  34.7% yield
theoretical yield
9.87 g
