Chem 104 Exam #4
Chapt 5-6 Winter 2006
Potentially Useful Information: Mercury (Hg) can be +1 or +2,
Lead (Pb or Plumbum) can be +2 or +4 Q (heat) = cmT
specific heats: water = 4.184 J/g C
lead = 0.13 J/g C
a.w. (g/mole): He = 4.003, O = 16.00, F = 19.00, H = 1.008
C= 12.01
0 C = 273.15 K, R = 0.08206 L atm/mole K
Dalton’s Law of partial pressures: pressure fraction = mole fraction
Name_______________________
Chem 104 Exam #4
Chapt 5-6
Winter 2006
Show all work, including any unit conversions for full credit, where applicable. Use
correct number of sig figs.
1. Balance the eqns below. (4 pts each)
2 H3PO3
+
1 Cl2 +
2 NaBr 
1 P4O10
1 CaCO3
3 Mg(OH)2
+
+

6 H2O +
2 NaCl +
1 Br2
4 H2O

2 HCl 
1 CaCl2
1 Mg3(PO3)2
2 H4P2O7
+
1 H2O +
1 CO2
2. A sample of Helium has a mass of 14.56 g. It is in a 19.00 L container at a
temperature of 14.0C. What is the pressure inside the container?
PV = nRT;
P= nRT/V
T = 14.0 + 273.15 = 287.2 K (have to round off
because moving from addition to multiplication).
P = nRT/V = (14.56g x 1 mole/4.003 g) x 0.08206 L atm/mole K x 287.2 K
19.00L
= 4.512 atm
3. Examine the reactions below. What are the redox numbers for each element in
each molecule? Which atom is oxidized and which reduced? Which atoms are
the oxidizing and reducing agents?
4 Fe + 3 O2

2 Fe2O3
ox #:
0
0
ea. +3 -2
+6 -6
Iron is oxidized, reducing agent
oxygen is reduced, oxidizing agent
CH4 +
2 O2 
ox # -4 +1
0
carbon is oxidized, reducing agent
CO2 +
2 H2O
+4 -2
+1 -2
oxygen is reduced, oxidizing agent
4. 136.4 g of oxygen gas, 14.9 moles of Argon gas, and 149.3 g of fluorine gas are
mixed together in a tube. If the total pressure is 1.470 atm, what is the pressure of
each individual gas?
Daltons law sez pressure fraction = mole fraction, so find mole fraction
O2 moles: 136.4 g x 1 mole/32.00 g
F2 moles: 149.3 g x 1 mole/38.00 g
Ar moles:
= 4.2625 moles
= 3.92895 moles
= 14.9 moles
23.0914… moles or 23.1 to correct sig figs
O2 fraction: 4.2625 moles/23.1 moles
F2 fraction: 3.92895 moles/ 23.1 moles
Ar fraction: 14.9 moles/23.1 moles
x 1.470 atm = 0.271 atm
x 1.470 atm = 0.250 atm
x 1.470 atm = 0.948 atm
5. 68.06 grams of sucrose and 172.4 grams of oxygen are mixed together to make
carbon dioxide and water. Which is the limiting reagent? How many grams of
water and carbon dioxide are produced?
C6H12O6 +
6 C = 72.06
12 H= 12.096
6 O = 96.00
180.16 g/mole
6 O2 
6 H2O +
C = 12.01
2O= 32.00
44.01 g/mole
6 CO2
O = 16.00
2 H= 2.016
18.02 g/mole
68.06 g x (1mole/180.16 g) x 6 CO2/1 C6H12O6 x 44.01 g/mole = 99.76 g
172.4 g x 1mole/32.00 g x 6 CO2/6 O2
x 44.01 g/mole = 237.1 g CO2
Sucrose is limiting, 99.76 g CO2 formed
68.06 g x 1mole/180.16 g x 6 H2O / 1 C6H12O6 x 18.02 g mole = 40.85 g H2O
6. A sample of gas is originally at STP (1 atm and 0 C) in a steel drum. It is heated
to 451 C. What is the new pressure?
First, convert temperature to Kelvin. K = C+273 = 451 + 273 = 724 K
P1V1
n1T1
=
P1/T1 =
P2 =
P2V2
n2T2
but Volume and #moles doesn’t
change, so V1=V2, and n1=n2
P2/T2 so,
P2
=
P1V
nT1
=
P2V
nT2
T1 x P2/T2
724 K x 1.00 atm/ 273.15 K = 2.65 atm
7. Write a total and net ionic eqn for the following reaction.
NH4Cl (aq) +
Pb(NO3)2 (aq) 
PbCl2 (s)+
NH4NO3 (aq)
First, balance the eqn: 2 NH4Cl (aq) + Pb(NO3)2 (aq)  PbCl2 (s)+ 2 NH4NO3 (aq)
Total:
2 NH4+ (aq) + 2 Cl- (aq) + Pb2+(aq) + 2 NO3- (aq)  PbCl2 (s)+ 2 NH4+ (aq) + 2 NO3- (aq)
Net: remove the ammonium and nitrate spectator ions
2 Cl- (aq) + Pb2+(aq)  PbCl2 (s)
Extra Credit:
A 25.6 g piece of lead at 149 C is dropped into 13.4 g of water at 20.3 C. What is the
temperature of the resulting solution?
The water heats up and the lead cools down. The heat lost by the lead is the same
magnitude but opposite in sign as the heat gained by the water.
-QPb = QH2O or
-cmT= cmT
where T = Tf –Ti
initial temperatures are the different for each, but the final T is the same for both.
-cPb mPb TPb = cH2O mH2O TH2O
- 0.13 J/g C x 25.6 g x (Tf – 149 C) = 4.184 J/g C x 13.4 g x (Tf – 20.3 C)
-3.328 J/C Tf +
495.872 C =
56.0656 J/C Tf –1138.13 C
59.3936 J/C Tf
=
1634.00368 J
Tf
=
28 C
You may notice that no attention was paid to sig figs until the final answer. That is okay
because it is only necessary to pay attention to sig figs when reporting an answer or when
switching back and forth between multiplication and addition (because the rules change
depending on which you are doing).