Chem 1151
Exam 2-Part 1A(green)
SHOW ALL WORK FOR CREDIT
Mar 25, 2008
NAME____________KEY___________
SEC_________
50 pts
R = 0.0821 (L-atm)/(mol-K) = 8.314 J/(mol-K)
NA = 6.022E+23/mol
1 J = 101.3 L-atm
1 atm = 760 mm Hg = 760 torr
EK = (3/2)RT
Vm = 22.4 L/mol at STP
urms = 3RT/M
1.
(7) If a 10.0 g sample of Zn(s) is treated with 75.0 mL of a 3.60M
aqueous HCl solution, what volume of hydrogen gas will be produced
at STP?
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
mass-Zn  mol-Zn  mol HCl needed
compare this with mol HCl on hand = M x V
[10.0g/(65.39g/mol)] (2 mol HCl/1 mol Zn) = 0.306 mol HCl needed
0.075L x 3.60 M = 0.27 mol HCl on hand, so HCl = LR
Mol HCl on hand  mol hydrogen produced  V = nRT/P (ideal gas law)
V = [0.27 mol HCl (1 mol H2/2 mol HCl)] [0.0821 x 273]/1 atm = 3.03 L
2.
(7) A 2.74 g sample of a colorless liquid was vaporized in a 250 mL
flask at 121 oC and 786 torr. Its empirical formula is C8H18. What is the
molar mass of this substance? What is its molecular formula?
Convert mL to L, torr to atm and Celsius to K
M = dRT/P = (2.74 g/0.250L) [0.0821 x 394]/[786/760)] = 342.8 g/mol
EF mass = 114.18
342.8/114.18 = 3 so MF = C24H54
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3.(7) Consider this neutralization reaction:
Ca(OH)2(aq) + 2HNO3 (aq)  2H2O(l) + Ca(NO3)2(aq).
a. (1) Calculate the molarity of hydroxide ions in a 0.60 M Ca(OH)2
solution.
0.60M base (2 mol OH-/1 mol base) = 1.2M OHb. (2) How many moles of hydrogen ions are present in 40.0 mL of a
1.25M HNO3 solution?
# mol = M x V = 0.0400L x 1.25M = 0.050 mol H+
c. (4) How many mL of 0.60 M Ca(OH)2 are required to neutralize 40.0
mL of a 1.25M HNO3 solution?
M and V of acid  # mol acid  # mol base  volume of base
(0.0400L x 1.25M) (1 mol base/2 mol acid)/ 0.60M = 41.7 mL base
4.
(7)
Consider this balanced chemical equation
NH4NO3(s)  2H2O(g) + N2 (g)
This reaction has a reaction yield of 40%. If 25.0 L nitrogen was collected
at 1.00 atm at 300K, how many moles of ammonium nitrate decomposed?
V, T, P of N2  mol N2 = PV/nRT  mol NH4NO3  adjust for 40%
yield
Based on chemical eqn, 1 mol NH4NO3) decomposed to give 1 mol N2
(25.0L x 1.00 atm)/[(0.0821 x 300K) (1 mol N2 /1 mol NH4NO3)]/0.40 = 2.54
mol NH4NO3
2
5. (10) Balance this redox reaction in base by the half-reaction method.
SHOW ALL STEPS FOR CREDIT.
OX# +3, -2, +1 +2, -2, +1
+4, -2, +1
Bi(OH)3 + Sn(OH)3 
Sn(OH)62- + Bi
Oxidation ½ Rxn:
[Sn(OH)3- + 3H2O  Sn(OH)62- + 3H+ + 2e-]3
Reduction ½ Rxn:
[Bi(OH)3 + 3H+ + 3e-  3H2O + Bi]2
Redox Rxn
3Sn(OH)3- + 9H2O + 2Bi(OH)3 + 6H+ + 6e- 
3Sn(OH)62- + 9H+ + 6e- + 6H2O + 2Bi
3Sn(OH)3- + 3H2O + 2Bi(OH)3 + 3OH- 
+2Bi + 3H+ + 3OH-
3Sn(OH)62- +
3Sn(OH)3- + 2Bi(OH)3 + 3OH-  3Sn(OH)62- + +2Bi
Check
R
3
18
18
2
-6
P
3
18
18
2
-6
Sn
O
H
Bi
ch
a.
What is the oxidizing agent?_____ Bi(OH)3 ___
b.
What is oxidized?______Sn___________
c. How many electrons does each Sn atom gain or lose? Sn loses 2
electrons
3
6.(12) a.
Calculate the volume of a 2.75 mol sample of CO gas at STP.
2.75 mol x 22.4 L/mol = 61.6 L
b.
Calculate the temperature of F2 that has the same root mean
square speed that CO2 has at 300K.
urms (F2) = 3RT/M = urms (CO2) = 3RT’/M’
M = molar mass of F2 ; M’ = molar mass of CO2; T’ = 300K
Solve for T of fluorine
T/M = T/0.038 = T’/M’ = 300/0.044  T = 259K
c.
If the temperature of CO gas is doubled while holding the
volume and number of moles of gas constant, the pressure will
1.
stay the same
2.
**double V = nRT/P  T
3.
be reduced by 50%
4.
increase four-fold
5.
not enough information
d.
At 300K, what is the ratio of the effusion rates of CO to CO2?
Effusion rate is proportional to urms which is proportional to 1/M.
Rate (CO) = 0.044
Rate (CO2 ) = 0.028 = 1.25
Extra Credit (2)
Hydrogen gas is collected over water at 19 oC and a total pressure of 769
torr. If 156 mL of gas is collected, how many moles of hydrogen were
collected? The vapor pressure of water at this temperature is 16.5 torr.
Calculate partial pressure of H2. Then use ideal gas law to find ni = PiV/RT
where i = H2
P H2 = PT – P H2O = 769-16.5 = 754.5 torr
ni = [754.5/760] 0.156L/[0.0821 x 292K) = 0.0064mol H2
4