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Practice Exam for Exam 2
1. Potassium permanganate solution reacts with sodium oxalate solution in the presence of sulfuric acid
in an oxidation-reduction reaction. Two of the products of this reaction are the manganese(II) ion
and carbon dioxide.
a. Write a balanced complete chemical equation for this reaction.
16 H2SO4 + 2 KMnO4 + 5 Na2C2O4  2 Mn(HSO4)2 + 10 CO2 + 2 KHSO4 + 10 NaHSO4 + 8 H2O
b. If 25.43 mL of 0.02532 M potassium permanganate solution reacts with 14.64 mL of 0.09383 M
sodium oxalate solution, how many liters of carbon dioxide gas are produced at 749.2 mmHg and
23.5°C?
0.02532 mol KMnO4 10 mol CO2
?L CO2 =
25.43 mL KMnO4 ×
×
1000 mL KMnO4
2 mol KMnO4
×
?L CO2 = 14.64 mL Na 2C2O4 ×
( 62.364 L mmHg mol
−1
)
K −1 ( 296.7 K )
( 749.2 mmHg )
0.07950 L CO2
=
0.09383 mol Na 2C2O4
10 mol CO2
×
1000 mL Na 2C2O4
5 mol Na 2C2O4
×
( 62.364 L mmHg mol
−1
)
K −1 ( 296.7 K )
( 749.2 mmHg )
0.06785 L CO2
=
0.06785 L of CO2 will be produced.
2. Balance the following oxidation-reduction reactions. Give the molecular equation if possible.
a. Au + HNO3  NO2 + Au+
0
+1 +5 -2
+4 -2
+1
H+ + Au + HNO3  NO2 + Au+ + H2O
Au + 2 HNO3  NO2 + AuNO3 + H2O
b. ClO- + CrO42-  Cr(OH)3 + ClO4-
+1 -2
+6 -2
+3 -2 +1
+7 -2
5 H2O + ClO- + 2 CrO42-  2 Cr(OH)3 + ClO4- + 4 OH-
3.
A researcher determines the molar mass of a gas by using the time it takes for 1.00 mole of the gas
to effuse compared to the amount of time it takes for 1.00 mol of argon to effuse at the same
temperature in the same apparatus. She measures the time for argon to effuse as 6.34 minutes.
1.00 mole of the unknown gas takes 8.25 minutes to effuse.
a. What is the density, in g L-1, of the unknown gas at STP?
t unk
=
t Ar
2
2
t 
Munk
 8.25 min 
−1
−1
⇒ Munk=  unk  M Ar= 
 ( 39.948 g mol )= 67.6 g mol
M Ar
 6.34 min 
 t Ar 
( 67.6 g mol ) (1.00 atm) =
( 0.082058 L atm mol K ) (273.15 K )
−1
MP
=
d =
RT
−1
3.02 g L−1
−1
b. At what Kelvin temperature would the unknown gas have an rms speed of 456.2 m s-1?
u=
rms
2
Murms
T
⇒=
=
3R
M
3RT
( 67.6 × 10 kg mol )( 456.2 m s =
)
3 ( 8.314 J mol K )
−3
−1 2
−1
−1
−1
564 K
4, Use the data in the table provided to calculate the amount of energy, in kJ, produced or consumed
when 100.00 grams of each of the reactants below is converted into products.
4 NH3 + 5 O2  4 NO + 6 H2O
=
∆H
∑
products
n∆H f −
∑
reactants
n∆H f
 4 mol NO   90.29 kJ   6 mol H2O   −241.8 kJ    4 mol NH3   −45.90 kJ   5 mol O2   0.00 kJ  
=

+

  − 




+
 mol rxn   mol NO   mol rxn   mol H2O    mol rxn   mol NH3   mol rxn   mol O2  
= −906.04 kJ mol −1
1 mol NH3
1 mol rxn −906.04 kJ
100.0 g NH3 ×
×
×
=
−1.330 × 103 kJ
17.0305 g NH3 4 mol NH3
mol rxn
100.0 g O2 ×
1 mol O2
1 mol rxn −906.04 kJ
×
×
=
−566.3 kJ
31.9988 g O2 5 mol O2
mol rxn
566.3 kJ of energy will be produced by the mixture.
5. Hydrogen Sulfide, H2S, is a poisonous gas with the odor or rotten eggs. The reaction for the
formation of H2S from the elements is
H2 + 1/8 S8(rhombic)  H2S
Use Hess’s Law to obtain the enthalpy change for this reaction from the following enthalpy changes:
-1(H2S + 3/2 O2  H2O + SO2)
H2 + ½ O2  H2O
1/8 S8(rhombic) + O2  SO2
H = -518 kJ×-1
H = -242 kJ
H = -297 kJ
∆H =
−21 kJ
( −1)( −518 kJ) + (1)( −242 kJ) + (1)( −297 kJ) =
6. A mixture contains calcium carbonate, CaCO3, and magnesium carbonate, MgCO3. A sample of this
mixture weighing 7.85 g was reacted with excess hydrochloric acid. The reactions are
CaCO3 + 2 HCl  CaCl2 + H2O + CO2
MgCO3 + 2 HCl  MgCl2 + H2O + CO2
If the sample reacted completely and produced 1.94 L of carbon dioxide, CO2, at 25C and 785 mmHg,
what were the percentages of CaCO3 and MgCO3 in the mixture?
mCaCO3 + mMgCO3 =7.85 g ⇒ mCaCO3 =7.85 − mMgCO3
n=
CO2
PV
=
RT
( 785 mmHg )(1.94 L ) =
−1
K −1 ) ( 298 K )
( 62.364 L mmHg mol
(7.85 − m
)
 1 mol CaCO3   1 mol CO2 
 1 mol MgCO3   1 mol CO2 
0.08194
= mCaCO3 



 + mMgCO3 
 100.09 g CaCO3   1 mol CaCO3 
 84.314 g MgCO3   1 mol MgCO3 
mMgCO3
MgCO3
+
= 7.85 − mMgCO3 ( 84.314 ) + mMgCO3 (100.09 ) =( 0.08194 )(100.09 )( 84.314 ) =691.4906980244
100.09
84.314
691.4906980244 =
661.8649 − 84.314mMgCO3 + 100.09mMgCO3 =
661.8469 + 15.776mMgCO3
0.08194 =
((
)
)
691.4906980244 − 661.8469 =29.6438 =15.776mMgCO3
29.6438
= 1.88 g ⇒ mCaCO3 = 7.85 g − 1.88 g = 5.97 g
15.776
1.88 g
% MgCO3 =
× 100 = 23.95% and %CaCO3 = 100 − 23.95 = 76.05%
7.85 g
mMgCO3 =