Chem 103 Exam 2 Solutions Spring, 2009
1. (4 pts) Balance the following equation:
1 CaCN
2
(s) + 3 H
2
O(l)  1 CaCO
3
(s) + 2 NH
3
(g)
2. (4 pts) The very stable compound SF6 is made by burning sulfur in an atmosphere of fluorine, according to the equation
S
8
(s) + 24 F
2
(g)

8 SF
6
(g).
If you need to produce 2.50 moles of SF
6
, how many moles of S
8
and F
2
will you need to use? a. 0.125 moles of S
8
and 3.00 moles F
2. b. 1.00 moles of S
8
and 24.0 moles of F
2 c. 0.313 moles of S
8
and 7.50 moles F
2 d. 0.313 moles of S
8
and 3.0 moles of F
2 e. 0.125 moles of S
8
and 7.50 moles of F
2
f. none of these answers is correct
3. (4 pts) Which of the following compounds would be expected to be soluble in water? Mark all correct answers. a. Cu
3
(PO
4
)
2 b. CuBr
2 c. (NH
4
)
2
S d. Fe(OH)
3 e. K
2
SO
4
4. (2 pts) In the reaction I + ClO 
IO + Cl , what species is the oxidizing agent? ClO -
5. (2 pts) What is the oxidation number for arsenic (As) in the compound H
2
AsO
4
-
? +5
6. (2 pts) A white solid is either NaCl or Na
2
SO
4
. If an aqueous solution is prepared, which reagent will allow you to distinguish between these two compounds? a. H
2
SO
4 b. HCl c. Pb(NO
3
)
2 d. Cu(NO
3
)
2
e. HBr f . none of these
7. (2 pts) Equal masses of two substances, A and B, each absorb 25 J of energy. The temperature of A increases by 4 degrees and the temperature of B increases by 8 degrees. Which of the following statements is true? Mark all correct answers. a. The specific heat of A is double that of B b. The specific heat of B is double that of A. c. The specific heat of B is negative. d. The specific heat of A is negative. e. The specific heat of B is triple that of A.
8. (4 pts) Write the reaction that illustrates a formation reaction for acrylonitrile, C
3
H
3
N.
3 C(s) + 3/2H
2
(g) + ½ N
2
(g)

C
3
H
3
N

9. (3 pts) For a particular process, q = 30 kJ and w = -25 kJ. What conclusions can be drawn for the process? Mark all correct answers. a.

E = 55 kJ b.

E = -55 kJ c.

E = 5 kJ d.

E = -5 kJ e. the transfer of heat energy is from the system to the surroundings f.
the transfer of heat energy is from the surroundings to the system g. the system does work on the surroundings h. the surroundings does work on the system
10. (3 pts) Which of the following are strong electrolytes? Mark all correct answers. a. HCl b. HF c. NaNO
3 d. Pb(OH)
2 e. C
6
H
12
O
6 f. K
2
SO
4
11. (3 pts) What is the total concentration of ions (M, moles/L) in 0.100M Ba(NO
3
)
2
? a. 0.100M b. 0.200M c. 0.300M
d. 0.500M e. 0.0500M f. 0.0333M
12. (3 pts) What is the pH of 2.0 x 10
-3
M HCl? pH = 2.70
13. (3 pts) If the pH of a given solution is 10.24, what is the [H
+
]? 5.75 x 10-11 M
14. (3 pts) Define ‘strong acid’ and give two examples.
A strong acid is completely (100%) dissociated. Examples include HCl, HBr, HI, HNO
3
,
H
2
SO
4
, HClO
4

Part II. Problems
1. (10 pts) Disulfur dichloride (S
2
Cl
2
) can be prepared by the following reaction:
3SCl
2
( l ) + 4NaF( s ) → SF
4
( g ) + S
2
Cl
2
( l ) + 4NaCl( s )
(molar masses: SCl
2
102.96, NaF 41.98, SF
4
108.05, S
2
Cl
2
135.02, NaCl 58.44)
If 39.5 g of SCl
2 and 21.4 g of NaF are reacted a. What is the limiting reactant? moles SCl2 = 0.3836 moles moles NaF = 0.5098 moles
4 molNaF
Need
3 molSCl
2 or
1 .
33 molNaF
1 molSCl
; have
0 .
5098 molNaF
0 .
3836 molSCl
2 or
1 .
328 molNaF
1 mol
Therefore, NaF is the limiting reagent. b. What is the theoretical yield of S
2
Cl
2
? gS
2
Cl
2

0 .
5098 molNaF x
1 molS
2
Cl
2
4 molNaF x
135 .
02 molS gS
2
Cl
2
2
Cl
2

17 .
2 gS
2
Cl
2 c. If 12.6 g of S
2
Cl
2 are actually obtained from the reaction, what is the percent yield ?
% yield

12 .
6 g
17 .
2 g x 100

73 .
3 %
2. (6 pts) Write the products for each reaction, balance, then write the net ionic equation for each. a. 2 AgNO
3
+ Na
2
S

Ag
2
S(s) + 2NaNO
3
(aq)
Net ionic: 2 Ag + (aq) + S 2(aq)

Ag
2
S(s) b. Fe
2
(CO
3
)
3
(s) + 6 HNO
3
(aq)

2 Fe(NO
3
)
3
+ 3 H
2
O + 3 CO
2

Net ionic: Fe
2
(CO
3
)
3
(s) + 6H + (aq)

2 Fe 3+ (aq) + 3 H
2
O + 3 CO
2
3. (4 pts) What mass of NaCl (MW 58.5 g/mol) should be measured out to prepare 400.0 mL of
0.125M NaOH? g

0 .
125 mol x
1 L
0 .
4000 L x
58 .
5 g
1 mol

2 .
93 g
4. (4 pts) A solution was prepared by pipeting 20.0 mL from a stock solution and diluting it to
150.0 mL. The molarity of the final solution was 0.247 M. What was the molarity of the original stock solution?
M i
V i
= M f
V f
:
M i

M f
V f 
( 0 .
247 M )( 150 .
0 mL

1 .
85 M
20 .
0 ml V i
5. (6 pts) Vitamin C has the formula C
6
H
8
O
6
(MW 176.2 g/mol) You can determine the amount of Vitamin C in a sample by titration with a solution of bromine, Br
2
, according to the equation:
C
6
H
8
O
6
(aq) + Br
2
(aq)

2 HBr(aq) + C
6
H
6
O
6
(aq)
If a 2.00-g vitamin tablet required 27.85 mL of a 0.102M Br
2
solution to the equivalence point, what is the mass percent of Vitamin C in the sample? g

0 .
102 mol
L x
0 .
02785 L x
176 .
2 g mol

0 .
501 g mass percent = 0.501 g/2.00 g x 100 = 25.1%
6. (6 pts) Calcium carbide (CaC
2
) is manufactured by the reaction of CaO with C at high temperatures. If 10.0 g of CaO is allowed to react with excess carbon, what quantity of heat is involved?
CaO(s) + 3C(s)

CaC
2
(s) + CO(g)

H rxn
= +464.8 kJ
Moles CaO = 10.0 g x 1 mol/56.08 g = 0.1783 moles kJ = 464.8 kJ/mol x 0.1783 moles = 82.9 kJ of heat involved

7. (8 pts) When 50.0 mL of 0.200 M KOH is mixed with 25.0 mL of 0.400 M HBr in a coffeecup calorimeter, the temperature of the solution rises from 22.50 to 24.28 °C. Assuming the densities of the solutions are all 1.00 g/ml and the heat capacities are all 4.184 J/g
⋅ °C, what is the enthalpy of the reaction per mole of KOH?
Mass of solution = 75.0 mL or 75.0 g (since density of water is 1.00 g/mL)
Change in temperature is 24.28 – 22.50 = 1.78 o C q soln
=
4 .
184 J g o
C x
75 .
0 g x
1 .
78 o
C

558 .
56 J q rxn
= -q rxn
= - 558.56 J moles KOH = 0.0500 L x 0.200 mol/L = 0.0100 mol KOH

H = - 558.56 J/0.0100 mol = -5.58 x 10 4 kJ/mol
8. (6 pts) Given the enthalpies of the following reactions:
(Eqn 1) 3C(s) + 3H
2
(g)  C
3
H
6
(g)
(Eqn 2) C(s) + O
2
(g)  CO
2
(g)
(Eqn 3) H
2
(g) + ½ O
2
(g)

H
2
O(l)



H = 53.3 kJ
H = -394 kJ
H = - 286 kJ
Calculate the enthalpy of reaction for the combustion to propane:
C
3
H
6
(g) + 9/2O
2
(g)

3CO
2
(g) + 3 H
2
O(l)

H = ?
Reverse Eqn 1:
Multiply Eqn 2 by 3
Multiply Eqn 3 by 3

H = - 53.3 kJ

H = - 1182 kJ

H = - 858 kJ

H rxn
= - 2093 kJ

9. (8 pts) The enthalpy change for the oxidation of styrene, C
8
H
8
, is measured by calorimetry:
C
8
H
8
+ 10 O
2
(g)

8 CO
2
(g) + 4 H
2
O(l)

H rxn
= -4395.0 kJ
Given the following standard enthalpies of formation, calculate the molar enthalpy of formation
(

H o f
) of styrene:

H o f
CO
2
(g) = -394 kJ

H o f
H
2
O(l) = -286 kJ

H rxn
  
H prod
  
H reactabts

H rxn

[8

H f
CO
2
+ 4

H f
H
2
O] – [

H f
C
8
H
8
]
-4395.0 kJ = [8(-394) + 4(-286)] - [

H f
C
8
H
8
]
-4395.0 = -3152 – 1144 - [

H f
C
8
H
8
]
-99 kJ = - [

H f
C
8
H
8
]

H f
C
8
H
8
= + 99 kJ
Extra Credit: (5 pts - All or nothing): The ice diet involves eating ice to lose weight. Calculate the amount of energy (in Joules) your body would need to expend to convert 1 gram of ice at 0 o
C to body temperature (37 o
C). To lose one pound of weight, you need to expend 3500 kcal.
How many pounds of ice would you need to eat in order to lose one pound? 1 cal = 4.184 J

H fus
= 333 J/g
C water
= 4.184 J/g o

H vap
= 2256 J/g
C C ice
= 2.06 Kg o
C
C vapor
= 1.92 J/g o C
Heat needed to raise melt ice and raise temperature to 37oC:
(333 J/g x 1 g) + (4.184 J/g o C x 1 g x 37 o C) = 488 J or 116 cal needed to heat 1 g
Pounds of ice needed = 3.5 x 106 cal
1 g ice requires 116 cal; x g requires 3.5 x 106 cal; x = 30172 g or 66 lbs of ice.