Name:___________________________
CHEM&141- EXAM II – F2008 – CHAPTERS 4-6 - 102 POINTS
1. If a reaction has a theoretical yield of 2.56 grams and a percent yield of 22.8, what
is its actual yield in grams? (4 points)
a. 0.584 g
b. 0.112 g
c. 8.91 g
d. None of the above
2. In the following reaction, which is oxidized and which is reduced? (4 points)
2NO3−(aq) + 12H+(aq) → N2(g) + 6H2O(l)
a.
b.
c.
d.
e.
NO3− is reduced and H+ is oxidized
NO3− is oxidized and H+ is reduced
NO3− is oxidized and H+ is oxidized
NO3− is reduced and H+ is reduced
This is not a redox reaction
For the following:
1) Clearly state the molecular and net ionic equations for each (include
phase labels!) If there is no reaction write “NR” (8 points – 24 points total)
2) Circle the type of reaction. For non-reactions just circle No Reaction
(NR). (4 points – 12 points total)
3. A solution of lead (II) permanganate reacts with a solution of sodium iodide to
yield a precipitate.
Acid-Base
Single Displacement
Decomposition
Redox
Double Displacement
No Reaction
Molecular equation:
Pb(MnO4)2(aq) + 2NaI(aq)  PbI2(s) + 2NaMnO4(aq)
Net-ionic equation:
Pb2+(aq) + 2I-(aq)  PbI2(s)
________/20 points total
ALIABADI
CHEM&141 EXAM 2 Fall2008
Page 1 of 5
4. Calcium metal is placed in water yielding insoluble calcium hydroxide and
hydrogen gas.
Acid-Base
Decomposition
Double Displacement
Single Displacement
Redox
No Reaction
Molecular equation:
Ca(s) + 2H2O(l)  Ca(OH)2(s) + H2(g)
Net-ionic equation:
Ditto
5. Calculate Hrxn, in kJ, for the following reaction: (10 points)
½ N2(g) + ½ O2(g)  NO(g)
Given:
N2(g) + 3H2(g)  2NH3(g); H = -91.8 kJ
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g); H = -906.2 kJ
H2(g) + ½ O2(g)  H2O(g); H = -241.8 kJ
2N2(g) + 6H2(g)  4NH3(g); H = -91.8 kJ x 2 = -183.6 kJ
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g); H = -906.2 kJ
6H2O(g)  6H2(g) + 3O2(g); H = +241.8 kJ x 6
2N2(g) + 2O2(g)  4NO(g); 361.0 kJ
Divide by 4
½ N2(g) + ½ O2(g)  NO(g); 90.3 kJ
_________/22 points total
ALIABADI
CHEM&141 EXAM 2 Fall2008
Page 2 of 5
6. 0.149 grams of quicksilver is placed in a coffee-cup calorimeter and then 125.0 mL
of 1.00 M HCl is added. The reaction that occurs is:
Hg(l) + 2HCl(aq)  HgCl2(aq) + H2(g)
The temperature of the solution increases from 25.89°C to 27.09°C. What is the
enthalpy change for the reaction per mole of Hg? Assume that the specific heat
capacity of the solution is 4.184 J/g°C and the density of the HCl solution is 1.00
g/mL. (MW of Hg = 200.59 g/mol) (10 points)
qsoln  mC T  (1.00 g
mL
125.0mL)  (4.184 J
g C
)(1.25 C )  628 J
qsoln  qrxn  628 J
H rxn 
628 J 200.59 g

 8.45 105 J
mol Hg
0.149 g mol Hg
7. Hydrated nickel (II) chloride is a beautiful, green, crystalline compound. If 0.235g
of this hydrate is heated, it will leave 0.128g of the salt behind. What is the formula of
the hydrated compound? (MW of NiCl2 = 129.5988 g/mol & MW of H2O = 18.0153
g/mol) (10 points)
0.128 g NiCl2 
1mol NiCl2
 9.88 104 mol NiCl2
129.5988g NiCl2
0.235 g  0.128 g  0.107g H 2O
0.107g H 2O 
1mol H 2 O
 5.94 103 mol H 2O
18.0153 g H 2 O
9.88 104 mol
 6.01  6
5.94 103 mol
Thus, chemical formula = NiCl2  6H 2 O
______/20 points total
ALIABADI
CHEM&141 EXAM 2 Fall2008
Page 3 of 5
8. 0.050 L of a solution of 1.03 M NiCl2 is mixed with 0.065 L of a solution of 1.13
M Hg2(NO3)2. What is the mass, in grams, of the precipitate produced? Hg2Cl2 =
472.09 g/mol, Ni(NO3)2 = 182.7033 g/mol. (10 points)
NiCl2(aq) + Hg2(NO3)2(aq)  Ni(NO3)2(aq) + Hg2Cl2(s)
1.03mol NiCl2 1molHg 2Cl2 472.09 gHg 2Cl2
NiCl2 : 0.050 L 


 24 gHg 2Cl2
1L
1molNiCl2
1molHg 2Cl2
Hg 2 ( NO3 ) 2 : 0.065 L 
1.13mol NiCl2 1molHg 2Cl2 472.09 gHg 2Cl2


 35 gHg 2Cl2
1L
1molNiCl2
1molHg 2Cl2
Thus, mass of precipitate produced is 24g.
9. A self-contained underwater breathing apparatus uses canisters containing
potassium superoxide. The superoxide consumes the carbon dioxide exhaled by a
person and replaces it with oxygen according to the equation below:
4KO2(s) + 2CO2(g)  2K2CO3(s) + 3O2(g)
What mass of the superoxide, in grams, is required to react with 8.90 L of carbon
dioxide at 22.0 °C and 767 mm Hg, where 760 mmHg = 1.00 atm? (MW of KO2 =
71.0971 g/mol & R = 0.08206 Latm/molK) (10 points)
PV  nRT
1atm
)  8.90 L
PV
760mmHg
n

 0.371moles CO 2
L  atm
RT
(0.08206
)(295.2 K )
mol  K
4 mol KO 2 71.0971g KO 2
0.371moles CO 2 

 52.8g superoxide
2 mol CO 2
1mol KO 2
(767mmHg 
_____________/20 points total
ALIABADI
CHEM&141 EXAM 2 Fall2008
Page 4 of 5
10. Eugenol is the major component in oil of cloves. It has a molar mass of 164.2
g/mol and is 73.14% C, 7.37% hydrogen, and 19.49% oxygen. What are the empirical
and molecular formulae of eugenol? (10 points)
1mol
73.14 gC 
 6.089molC  1.218mol  4.999C  5C
12.011g
1mol
7.37 gH 
 7.31molH  1.218mol  6 H
1.00794 g
1mol
19.49 gO 
 1.218molH  1.218mol  1H
15.9994 g
Thus, empirical formula = C5 H 6O
empirical formula mass = 82.102 g
mol
164.2 g
mol  2.000
g
82.102
mol
And that makes the molecular formula = C10 H12 O 2
n
11. A small bubble rises from the bottom of a lake, where the temperature and
pressure are 8°C and 6.4 atm, to the water’s surface, where the temperature is 25° and
the pressure is 1.0 atm. Calculate the final volume of the bubble if its initial volume
was 2.1mL. (10 points)
PV
PV
1 1
 2 2
T1
T2
(6.4atm)(2.1mL) (1.0atm)V2

(281K )
(298 K )
V2  14mL
___________/20 points total
ALIABADI
CHEM&141 EXAM 2 Fall2008
Page 5 of 5