CHEM 171
EXAMINATION 2
November 6, 2008
Dr. Kimberly M. Broekemeier
NAME:___________________Key____________________________________
KE = NA½ mµ2 = 3/2 RT
IA
II A
III B
IV B
VB
22.4 L/mol (at STP)
VI B VII B
IB
VIII
II B
III A
IV A
VA
VI A
1
VII A inert
gase
s
2
H
He
1.008
4.003
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
6.941 9.012
10.81 12.01 14.01 16.00 19.00 20.17
11
12
13
14
15
16
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
22.98 24.31
26.98 28.09 30.97 32.06 35.45 39.95
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39.10 40.08 44.96 47.90 50.94 52.00 54.94 55.85 58.93 58.71 63.55 65.37 69.72 72.59 74.92 78.96 79.90 83.80
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
85.46 87.62 88.91 91.22 92.91 95.94 98.91 101.0 102.9 106.4 107.8 112.4 114.8 118.6 121.8 127.6 126.9 131.3
55
56
Cs
Ba
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
*
132.9 137.3 138.9 178.4 180.9
87
88
89
104
105
Fr
Ra
(223) 226.0
*
Ac
Rf
Db
(227)
(261)
(262)
183.8
106
Sg
186.2 190.2 192.2 195.0 196.9 200.5 204.3 207.2 208.9 (210)
107
108
109
110
111
112
113
114
Bh
Hs
(263) (262) (265)
(210) (222)
Mt
(266) (269) (272) (277)
?
(289)
58
59
60
61
62
63
64
65
66
67
68
69
70
71
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
140.1 140.9 144.2 (147) 150.4 151.9
90
91
92
93
94
95
Th
Pa
U
Np
Pu
232.0 (231) 238.0 (237) (242)
157.3 158.9 162.5 164.9 167.3 168.9 173.0 174.9
96
97
98
99
100
101
102
103
Am
Cm
Bk
(243)
(247)
(247)
Cf
Es
Fm
(251) (254) (253)
Md
No
Lw
(256) (254) (257)
Part I: Short Answer (46 points total)
1) Give the oxidation number of:
P in PO43- ___+5____
C in C2O42- ___+3_____
I in HIO4 ___+7____
2) The proton concentration of a solution is 5.6 x 10-5 M. pH = -log(5.6 x 10-5)
a) The pH of this solution is ___4.23________
b) This solution is (
acidic
neutral
basic
)
3) A container contains 0.75 atm He and 0.44 atm Ar. The total number of moles in the container is 0.80.
How many moles of He are in the container?
________________
ΧHe x ntot = moles He
(0.75 atm/(0.75 atm x 0.44 atm)) x 0.80 mole = 0.50 moles He
4) Circle the formulas of the substances which are insoluble in aqueous solution:
CuI
Rb2SO4
CaS
MgCO3
Mn(NO3)2
5) Write formulas of probable products when the substances below are mixed together. If no net reaction
occurs, write NONE in the space. You do not have to balance the equation. Include state/phase.
a. HCl(aq) + Ca(OH)2(s) :
____CaCl2(aq) + H2O (l)__________
b. KClO4(aq) + Ca(CH3COO)2(aq) :
__None______________________________
c. K3PO4(aq) + CaCl2(aq):
____Ca2(PO4)3(s) + KCl (aq) _________________
d. CH3NH2(aq) + HI(aq):
____CH3NH3+(aq) + I-(aq)___________________
6) Write the net ionic equation for the following chemical reaction.
3 SrBr2(aq) + 2 Li3PO4(aq) → Sr3(PO4)2(s) + 6 LiBr(aq)
3 Sr2+ (aq) + 2 PO43- (aq) → Sr3(PO4)2 (s)
7) Balance the following redox half-reaction taking place in basic medium:
MnO4-(aq) → MnO2(s)
3e- + 2 H2O + MnO4- → MnO2 + 4 OH-
8) A particular chemical reaction is endothermic.
a) This means that the enthalpy of the products is (
greater than
less than
the enthalpy of the reactants.
b) This means that if the reaction is carried out in a closed flask, the flask will
(
increase
stay the same
decrease
) in temperature.
equal to
)
9) Balance the following redox reaction. Provide the overall balanced ionic equation. Do this by the method of half
reactions. The reaction takes place in acidic medium. Also, fill in the blanks with the formula of the species which
is acting as the oxidizing and reducing agent.
Oxidizing agent: ______NO3-__________
Reducing agent: _______Zn(s)________
Zn(s) + NO3-(aq) → Zn2+(aq) + N2O(g)
Oxidation half reaction: Zn → Zn2+ + 2eReduction half reaction: 8e- + 10H+ + 2NO3- → N2O + 5 H2O
Multiply oxidation half reaction by 4 to get
4 Zn(s) + 10 H+(aq) + 2 NO3- (aq) → 4 Zn2+ (aq) + N2O (g) + 5 H2O (l)
10) Ammonia and oxygen were placed in a rigid sealed container at constant temperature and allowed to react
according to the reaction below. The pressure due to ammonia was 335 torr. The pressure due to oxygen was
412 torr. Pressure is proportional to moles and can be used as is.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
335 torr NH3 x 4 mol NO/4 mol NH3 = 335 torr NO
412 torr O2 x 4 mol NO/5 mol O2 = 329 torr NO
O2 is limiting
a) Circle the formula of the limiting reactant.
b) After the reaction goes to completion, the total number of particles is ____ when the reaction began.
(
greater than
less than
the same as
) 9 moles to 10 moles
c) If T and V of the container remain constant, the pressure will be ____ as when the reaction began.
(
greater than
less than
the same as
) pressure is proportional to moles
11) In the space below, draw two sets of x,y axes. Label the axes and the plots clearly.
a) In graph a, draw a plot of the volume of a gas sample versus the number of moles present at constant
temperature and pressure.
b) In space b, draw a plot of the distribution of velocities of N2 molecules at some temperature T1 and
show (with a second plot) the effect of increasing to temperature to a higher, T2, value.
a.
b.
a) V is directly proportional to moles. With y axis Volume and x axis moles, you would have a straight line
going through 0,0.
b) y axis should be number of molecules and x axis should be velocity. At T1 there should be a curve that is
symmetrical and has a maximum at some velocity. At T2, there should also be a curve, with the maximum
at a higher velocity.
Part II. Multiple Choice (36 Points) For each of the following questions, circle the letter of the BEST response.
1) 1.36 g of K2S(s) is dissolved in enough water to prepare a solution which is 7.00 X 10-4 mol K+(aq)/L. What is
the volume of the solution? 1.36 g x mol/110.27 g x 2 mol K+/mol K2S = 2.47 x 10-2 mol K+
2.47 x 10-2 mole/x L = 7.00 x 10-4 M and x = 35.2 L
a. 0.476 mL
c. 38.6 mL
e. 52.5 mL
b. 35.2 L
f. 3.89 X 103 L
d. 150 L
2) In a titration of 25.0 mL phosphoric acid to K3PO4, 38.5 mL of 0.302 mol/L KOH(aq) titrant was required to reach
the equivalence point (according to the equation below). What is the molar concentration of the H3PO4 solution?
3 KOH(aq) + H3PO4(aq) → K3PO4(aq) + 3 H2O(l)
0.385 L x 0.302 mol KOH/L x 1 mol H3PO4/3 mol KOH = 0.003875 mol H3PO4 in 0.025 L
0.003875 mol/0.025 L = 0.155 M
a. 0.101 mol/L
c. 0.155 mol/L
e. 0.199 mol/L
b. 0.465 mol/L
d. 0.596 mol/L
f. 0.906 mol/L
3) A gas sample is placed in a 5.00 L container at 400 K and the pressure is determined to be 800 Torr. If both
the temperature and the volume are halved, what is the pressure under the new conditions?
P1V1/n1T1 = P2V2/n2T2, and n drops out, and T2 = ½ T1 and V2 = ½ V1 effect cancels
a. 800 Torr
c. 400 Torr
e. 200 Torr
b. 1200 Torr
d. 1600 Torr
f. 3200 Torr
4) Which one of the following gases has the slowest rate of diffusion at STP? The largest molar mass
a. N2
c. C2H6
e. SnH4
b. He
d. Kr
f. SO2
5) 300 Torr of CO(g) is mixed with 150 Torr of O2(g) in a sealed glass flask at 1000 K. The gases are allowed to
react according to the equation below. What is the final pressure (at 1000 K) after the reaction is complete?
2 CO(g) + O2(g) → 2 CO2(g)
Stoichiometric amounts and there is no reactant left, only product formed, use either reactant, not both.
300 Torr CO x 2 mol CO2/2 mol CO = 300 Torr CO2
a. 0 Torr
c. 0.45 Torr
e. 150 Torr
b. 300 Torr
d. 450 Torr
f. 750 Torr
6) An unknown gas took 1.33 times longer than argon to diffuse down a drift tube. What is the molar mass of this
unknown? Time is inversely proportional to velocity and Ar diffusion time x 1.33 = unk diffusion time
1.33 = (Munk/MAr)1/2
a. 5.48 g/mol
c. 22.6 g/mol
e. 30.0 g/mol
b. 53.1 g/mol
d. 70.7 g/mol
f. 2823 g/mol
7) A gas mixture of 1.60 g He and 0.9 g F2 occupies 15.0 L at 273 K, what is the partial pressure of helium?
a. 0.597 atm
c. 0.639 atm
e. 0.934 atm
b. 2.39 atm
d. 8.96 atm
f. 60.5 atm
Can simply use mass of He to get moles of He and calculate pressure due only to He.
1.60 g x mol/4.00 g = 0.4 mol
P = nRT/V = ((0.40 mol)(0.0821 L x atm/mol x K)(273K))/15.0 L
8) 0.562 g of a gas sample is placed in a 2.50 L flask at 25 oC. The pressure in the flask is measured as 28.4 Torr,
what is the molar mass of this substance? Calculate moles n = PV/RT = (0.0374 atm)(2.5 L)/(0.0821
Latm/molK)(298) = 0.00382 mol
0.562g/0.00382 mol = 147 g/mol
a. 262 g/mol
c. 147 g/mol
e. 58.8 g/mol
b. 262 g/mol
d. 19.6 g/mol
f. 12.6 g/mol
9) What is the density of nitrogen gas (N2) at STP? At STP 1 mol of any gas is 22.4 L; mass of 1 mol N2 is 28g
so 28.0 g/22.4 L = 1.25 g/L
a. 1.25 g/L
c. 1.16 g/L
e. 2.80 g/L
b. 0.625 g/L
d. 0.579 g/L
f. 1.40 g/L
10) How much heat energy (in kJ) is released by the complete combustion of 50.0 g methane (CH4)?
50.0 g x mol CH4 /16.024 g x 891kJ/mol CH4 = 2780 kJ
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
ΔH = -891 kJ
a. 285.8 kJ
c. 44550 kJ
e. 2777 kJ
b. 17.82 kJ
d. 891 kJ
f. 3000 kJ
11) In the lab, you titrated CoC2O4C2H2O with KMnO4(aq). This was done to determine:
a. mass % cobalt in salt
d. mass % oxalate ion in salt
b. % yield of salt synthesis
e. molar concentration of KMnO4(aq)
c. theoretical yield of salt
f. mass of oxide product
12) A student synthesized 1.43 g of CoC2O4C2H2O(s) (182.99 g/mol) using 3.19 g CoSO4C7H2O(s) (281.13 g/mol)
and 1.67 g H2C2O4C2H2O(s) (126.98 g/mol) as reactants. Given that the theoretical yield of product was
2.08 g, what is the percent yield for the synthesis? Simply use amount obtained 1.43 g
1.43/2.08 x 100
a. 80.3 %
c. 68.8 %
e. 45.5 %
b. 42.8 %
d. 36.1 %
f. 29.4 %
Part III: Problems (18 points total)
Complete three of the following four problems, and place an X through the problem you do not want graded. Show all
logic and work, using units. Report all results to three significant figures.
1) Calculate the volume of 0.120 M HCl(aq) required to react completely with 0.250 g of Ba(OH)2(s).
Always - balanced equation 2 HCl (aq) + Ba(OH)2 (aq) → BaCl2 (aq) + 2H2O (l)
Path: moles base
to moles acid to volume acid
0.250 g Ba(OH)2 x
mol Ba(OH)2
2 mol HCl
L
x
x
= 0.0243 L = 24.3 mL
171.32 g
mol Ba(OH)2
0.120 mol
2) 200 L TiCl4(g) at 5.00 atm and 300 C was reacted with 850 g H2O (l) and maximum product was formed. After
the reaction was complete, the system was adjusted to STP. What volume (L) of HCl(g) would be present at STP?
TiCl4(g) + 2 H2O(l) → TiO2(s) + 4 HCl(g)
Determine limiting reagent:
a) 850 g H2O x mol H2O/18.016 g x 4 mol HCl/2 mol H2O = 94.4 mol HCl
b) 200 L is amount TiCl4, use PV=nRT to convert to moles
n = PV/RT = (5 atm)(200L)/(0.0821 Latm/molK)(303K) = 40.2 mol TiCl4 x 4 mol HCl/1 mol TiCl4 = 160.6 mol HCl
H2O is limiting and 94.4 mol HCl will form, this will be at condition of STP
Use PV=nRT to calculate volume under STP.
V = nRT/P and (94.4 mol)(0.0821 Latm/molK)(273K)/1 atm = 2116 L = 2120 L
3) Calculate the root mean square velocity for molecular nitrogen at STP.
 rms =
3RT
=
M
3(8.314 J/molK)(273K)
= 493 m/s
0.028 kg/mol
4) Phosphorus trichloride reacts with water to form phosphorous acid and hydrochloric acid according to the
equation below. Molar masses, in g/mol, are given in parentheses below each compound. Calculate the kJ of
heat energy given off during this reaction, if 12.4 g of PCl3 was mixed with 4.31 g of H2O,
PCl3(l) +
(137.3 )
3 H2O(l)
(18.02 )
→
H3PO3(aq)
(81.99 )
+
3 HCl(aq)
(36.45 )
ΔH = -1140 kJ
Determine the limiting reagent:
12.4 g PCl3 x mol PCl3/137.3 g x 3 mol HCl/mol PCl3 = 0.27 mol HCl
4.31 g H2O x mol H2O/18.02 g x 3 mol HCl/3 mol H2O = 0.239 mol HCl
H2O is limiting. There is 0.239 mol of H2O consumed.
0.239 mol H2O x 1140 kJ/mol 3 mol H2O = 90.8 kJ released (-90.8 kJ)