Download 2011 Exam 2 Key

Prof. Thomas Greenbowe
Dr. Jesudoss Kingston
CHEM 177
Hour Exam II
October 3, 2011
This exam consists of 3 parts
on 8 pages (21 Questions)
Parts
Grading
Points
Page 3
21 pts
Score
24 pts
sheets only
Page 5
20 pts
Page 6
17 pts
Page 7
21 pts
TOTAL
103 pts
Recitation TA_______________
Recitation Section____________
on scantron
Page 4
Answer Key
Name______________________
________
________
TA Name
Abekoon
Anderson
Berry
Carlson
Chatterjee
Durfey
Everett
Fritzsching
Gupta
Hutchinson
Johnson
Khan
Klobukowski
Kumar, N.
Kumar, S.
Kwolek
Lampland
Maligal Ganesh
Marchuk
Pindwal
Reinig
Stender
Tomlinson
Van Zeeland
Walen
Wanninayake
Weinstein
Wiley
Zenner
Sections
21,57
11,30,46
39,62
55
29,42
26,63
8,60
47,54
25,33
6,48
40
37,43
12,44
20,28
1,2
5,9
38,41
13,18
45
31,59
15,56
19
7,10
36,52
3,22
4,64
53,58
24
23,27
Time
10, 12
2, 5, 3
11, 8
10
4, 1
3, 9
1, 5
4. 5
2, 9
11, 5
12
10, 1
3, 2
9, 4
8, 9
11, 1
11, 12
3, 5
2
8, 2
4, 11
8
12, 2
10, 3
10, 11
10, 12
4, 1
1
12, 3
NOTE: To receive full credit on problems, you must clearly show all work and your method of determining
the answer must be clear. The final answer must be reported to the correct number of significant figures and
have the correct units. Questions are written on both sides of each page. The last page contains useful
information and a periodic table; the last page should be removed and used for scratch paper and as a reference.
Do not put answers on the tear away page.
2
Please read the following instructions carefully before proceeding! Part I &II of your exam will be computer
graded. In order for the computer to identify who you are, it is important that you complete the information
section properly.
You must use a #2 pencil and completely fill in the appropriate circles on the RED computer scan sheet.
1.
To help you code the correct circles, first
write your last name, first name and
middle initial in the boxes (skip a space
between each). Then darken the circles
that match the letters in the box above it.
See the sample to the right.
2.
Write the middle nine digits of your ISU
identification number in the boxes A-I.
Do not skip any spaces. Below each
number, darken the circle that matches
this number. For example, 123456789.
See the sample at bottom right.
3.
Write your recitation section number in
the special code area, boxes K-L. Do not
skip any spaces. For example, if you are
in section 8 of Chem 177, write 08.
Again, darken the circle that matches the
number above it. See the sample at
bottom far right.
In Part I & II, select the one best answer for
each question. Place your answer on the
computer answer sheet by darkening the proper
circle for that question. Your computer scan
sheet will be your official answer sheet for Part
I & II.
All material (exam, answer sheet, scratch
paper) must be returned to your TA in order for
us to grade your exam.
Exam #2: Chem 177, Fall 2011
3
Part I: Multiple Choice: (3 pts each). The answer you fill in on your bubble sheet is the one that will count.
You should circle the answer on this sheet for your own reference.
1.
How many oxygen atoms are there in 10.0 grams of silver nitrate, AgNO3?
[molar mass of silver nitrate = 170.0 g]
a) 0.0588
2.
c) 1.06 × 1023
b) 0.176
d) 6.02 × 1023
e) 1.81 × 1024
The following diagram represents the reaction of A2 molecule (shaded spheres) with B2 Molecule
(unshaded spheres) to form AB3 molecules. Identify the limiting reactant and write a balanced chemical
equation for the reaction.
A
B
End
Start
a)
b)
c)
d)
e)
3.
A2 is the limiting reactant:
A2 is the limiting reactant:
B2 is the limiting reactant:
B2 is the limiting reactant:
B2 is the limiting reactant:
A + 3 B  AB3
A2 + 3 B2  2 AB3
A + 3 B  AB3
A2 + 3 B2  2 AB3
6A2 + 6 B2  4 AB3 + 4A2
Which one of the following is a strong electrolyte?
a) 1.0 M glucose
d) water
4.
b) HClO3
b) CuS
d) H2SO4
e) HCl
c) Pb(NO3)2
d) Al(OH)3
e) ZnCO3
Identify the major ionic species present in an aqueous solution of barium perchlorate, Ba(ClO4)2.
a) Ba2+, Cl–, O42–
d) Ba2+, ClO4–
7.
c) HBr
Which one of the following compounds will be water soluble?
a) BaSO4
6.
c) 1.0 M acetic acid
Which one of the following is a weak acid?
a) HNO2
5.
b) 0.50 M lithium chloride
e) 0.50 M ammonia
b) Ba, 2ClO4
e) Ba2+, ClO42–
c) Ba+, C4+, l–, O42–
The reaction between aqueous solutions of Iron(III) chloride, FeCl3 and barium hydroxide, Ba(OH)2 is
best described as ____________.
a) redox reaction
d) single displacement reaction
b) precipitation reaction
e) decomposition reaction
c) acid-base reaction
4
8.
When aqueous solutions of sodium phosphate Na3PO4(aq) and silver nitrate, AgNO3(aq) are mixed,
_____________.
a)
b)
c)
d)
e)
9.
there will be no reaction.
a precipitate of AgNa would form.
NO2 gas would be liberated.
Na+ and NO3 ions are the spectator ions and AgNO3 would precipitate.
Ag3PO4 will precipitate and Na+ and NO3 are the spectator ions.
What is the oxidation number of phosphorus atom in sodium dihydrogen phosphate, NaH2PO4?
a) –3
10.
b) –5
c) +7
d) +5
e) +3
d) H2
e) H+ in HCl
Which one of the following is a redox reaction?
a) HNO3(aq) + LiOH(aq)  LiNO3(aq) + H2O()
b) N2(g) + 3 H2(g)  2 NH3(g)
c) MgO(s) + H2O()  Mg(OH)2(aq)
d) 2 HClO3(aq) + PbS(s)  Pb(ClO3)2(aq) + H2S(g)
e) none of the above
11.
In the following redox reaction, what is being oxidized?
Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g)
a) Zn
b) HCl
c) ZnCl2
Part II: Multiple Choice: (4 pts each). The answer you fill in on your bubble sheet is the one that will count.
You should circle the answer on this sheet for your own reference.
12.
An unknown compound [Molar mass 130.0 g/mol] found to contain only carbon, nitrogen and hydrogen.
A 12.50 gram sample of this compound was analyzed to contain 73.85% carbon, 21.54% nitrogen and
4.61 % Hydrogen. What is the empirical formula of this compound?
a) C3H3N2
13.
c) C4H3N
d) C6H4N2
e) C8H6N2
How many grams of AgNO3 are needed to make 250. mL of a solution that is 0.135 M? [Molar mass of
AgNO3 = 170.0 g]
a) 0.0917 g
14.
b) C3H2N
b) 0.174 g
c) 5.74 g
d) 19.7 g
e) 91.7 g
What is the molar concentration of sodium ions in a 250.0 mL aqueous solution containing 16.4 g of
sodium phosphate, Na3PO4. [Molar mass of Na3PO4 =164 g]
a) 0.0012 M
b) 0.263 M
c) 0.400 M
d) 1.20 M
e) 1.40 M
5
15.
If the percent yield for the formation of HNO3 in the following
reaction is 75%, and 46.0 g of NO2(g) are consumed in the
reaction, how many grams of nitric acid, HNO3(aq) are produced?
NO2
H2O
HNO3
NO
3 NO2(g) + H2O(l)  2HNO3(aq) + NO(g)
a) 31.5 g
16.
b) 42.0 g
b) 0.50 M
e) 73.4g
c) 0.60 M
d) 0.70 M
e) 1.0 M
What is the concentration of HCl in the final solution when 65 mL of a 12 M HCl solution is diluted
with pure water to a final volume of 0.15 L?
a) 2.82 × 102 M
18.
d) 69.3 g
What is the molar concentration of chloride ions Cl in a solution made by combining 10.0 mL of
0.20 M NaCl and 10.0 mL of 0.40 M CaCl2? You may assume that volumes are additive.
a) 0.30 M
17.
c) 54.8 g
Molar masses (g/mol)
46.0
18.0
63.0
30.0
b) 5.2 M
d) 5.2 × 103 M
c) 28 M
e) 4.2 M
Box (1) represents a 1.0 mL solution of particles at a given concentration. Box (1) is then diluted to 5.0
mL as shown in box (2). 1.0 mL of this dilute solution was then transferred to each of other boxes.
1.0 mL
1.0 mL
(3)
(4)
1.0 mL
1.0 mL
(1)
(2)
(5)
(6)
Which one of the boxes among (3) – (6) represents a correct picture diagram of a 1.0 mL sample of that
diluted solution?
a) box (3)
19.
b) box (4)
c) box (5)
Calculate the theoretical yield of the precipitate when 25.0 mL of
0.50 M NaCl is mixed with 50.0 mL of 0.50 M AgNO3 solution.
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
a) 1.8 g
b) 0.0125 g
c) 0.013 g
d) box (6)
NaCl
NaNO3
AgNO3
AgCl
d) 3.6 g
e) none of the above
Molar masses (g/mol)
58.45
85.0
170.0
143.45
e) 0.56 g
6
Part-III: For full credit, show all your work legibly, include units, and report your answer to the correct
number of significant figures. No work shown = 0 points.
20.
(17 pts) Octane, C8H18 reacts with oxygen gas to form carbon
dioxide and water as shown below.
2 C8H18 (l) + 25 O2(g)  16 CO2 (l) + 18 H2O(l)
C8H18
O2
CO2
H2O
Molar masses (g/mol)
114.0
32.0
44.0
18.0
a) (4 pts) What is the theoretical yield (maximum calculated amount) of carbon dioxide that can form
when 10.00 grams of octane are heated with 10.00 grams of oxygen gas?
 1 mole C8H18   16 mole CO2   44.0 g CO2 

 = 30.9 g CO2 (3 Sig Fig)

114.0 g C8H18 2 mole C8H18 1 mole CO2
10.0 g C8H18 
1 mole O2 16 mole CO2  44.0 g CO2 


 = 8.80 g CO2 (Sig Fig 3)
 32.0 g O2   25 mole O2  1 mole CO2
10.0 g O2 
So, theoretical yield is 8.8 g of CO2
b) (2 pts) The limiting reagent is __Oxygen (O2)___. Explain.
10.0 g of Oxygen is able to produce only 8.8 g of CO2, although we have C8H18 enough to produce
30.9 g of CO2. The amount of oxygen is NOT enough to consume all of octane. It will run out first.
c) (4 pts) How many grams of the other reactant remains un-reacted?
Some octane will be left over. To Find left over Octane, we need to find how much is used up to
consume 10.0 g of O2.
1 mole O2 2 mole C8H18 114.0 g C8H18


 = 2.85 g C8H18 (Sig Fig 3)
 32.0 g O2   25 mole O2   1 mole C8H18 
10.0 g O2 
Therefore, mass of C8H18 left un-reacted = 10.0 – 2.85 = 7.2 g (2 Sig Fig)
d) (3 pts) If 6.55 g of CO2 are actually made in the laboratory under the above conditions, determine the
percent yield.
 actual yield 
6.55 g
Percent Yield = 
× 100 = 

 × 100 = 74 %
theoretical yield
 8.8 g 
e) (4 pts) A small volume of the starting material of the above reaction at the molecular level is shown in
the circle marked “start”. Complete the following picture diagram of what the atoms and molecules will
look like at the “end” of the reaction.
Start
10 C8H18
100 O2
End
64 CO2
72 H2O
2 C8H18
2 C8H18 (l) + 25 O2(g)  16 CO2 (l) + 18 H2O(l)
Multiply this by 4, to get 100 O2
8 C8H18 + 100 O2  64 CO2 + 72 H2O
Thus, there will be 64 CO2, 72 H2O and 2 C8H18 (unreacted)
7
21. (21 pts) Consider the reaction between dihydroxy fumaric acid and sodium hydroxide as shown below.
Dihydroxy fumaric acid is a diprotic acid.
a) (2 pts) Balance the following molecular equation.
2 NaOH(aq)
+ H2C4H2O6 (aq)  Na2C4H2O6 (aq) + 2 H2O(l)
b) (1 pt) Is the dihydroxy fumaric acid is a strong acid or a weak acid? Circle your choice.
Strong
Weak
c) (3 pts) Write the full ionic (or complete ionic) equation. (Be sure to include the phases)
2 Na+(aq) + 2 OH (aq) + H2C4H2O6(aq)  2 Na+(aq) + C4H2O62 (aq) + 2 H2O(l)
d) (3 pts) Write the net ionic equation. (Be sure to include the phases)
2 OH (aq) + H2C4H2O6(aq)  C4H2O62 (aq) + 2 H2O(l)
e) (4 pts) How many milliliters of 0.150 M sodium hydroxide need to completely neutralize 30.00 mL of
0.250 M dihydroxyfumaric acid (H2C4H2O6). What big idea are you using to solve this problem? Briefly
explain
BIG Idea: Since the acid is a diprotic acid, we need twice the number of moles of OH to completely
neutralize the acid [which can donate 2 H+ ions].
Moles of acid = 0.250 M 0.03000 L = 0.00750 mole
Now the balanced equation says that every 1 mole of acid, we need 2 moles of NaOH.
Moles of NaOH needed = 0.00750 mole H2C4H2O6 
2 mole NaOH 
= 0.0150 mole NaOH
 1 mole H2C4H2O6
Volume of NaOH needed = 
0.0150 mole NaOH
= 0.100 L = 100. mL (3 Sig Fig)
 0.150 M NaOH 
f) (2 pts) Is there a limiting reagent? Explain your answer in writing and justify with your numerical
answer from part-e.
There is no limiting reagent b’cos for neutralization reaction, number of moles of H+ from acid should be
equal to number of moles of OH from base. Thus, we have 0.00750 mole of acid (0.0150 mole of H+ and we
have 0.0150 mole of NaOH to neutralize.
g) (6 pts) A) In the first circle, a small volume of the starting material of the above reaction at the
molecular level is shown. This should represent the reactants as shown in your molecular equation.
Complete the picture diagram by writing enough number of NaOH. B) In the middle circle, write the
starting reagents and the number of each species, as represented in your full ionic equation in part-c.
C) In the last circle, write what species exist at the end of this acid-base neutralization reaction.
Remember to obey the Law of conservation of atoms.
10 H2C4H2O6
20 NaOH
20 Na+
20 OH
10 C4H2O62
20 Na+
20 H2O
Some additional Information:
In the middle circle:
Since H2C4H2O6 is a week acid, you can
write the dissociation of acid as shown
below:
H2C4H2O6  H+(aq) + HC2H2O6(aq)
Eventhough it is a diprotic acid, it will not
donate both protons that easily.
8
Avogadro’s number: 6.022  10
23
Pe rio d ic Ta b le o f t he Ele m e nt s
8A
18
1A
1
1
H
1 .0 1
2A
2
3A
13
4A
14
5A
15
6A
16
7A
17
2
He
4 .0 0
3
Li
6 .9 4
4
Be
9 .0 1
5
B
1 0 .8
6
C
1 2 .0
7
N
1 4 .0
8
O
1 6 .0
9
F
1 9 .0
11
Na
2 3 .0
12
Mg
2 4 .3
3B
3
4B
4
5B
5
6B
6
7B
7
8
9
10
1B
11
2B
12
14
Si
2 8 .1
15
P
3 1 .0
16
S
3 2 .1
17
Cl
3 5 .5
19
K
3 9 .1
20
Ca
4 0 .1
21
Sc
4 5 .0
22
Ti
4 7 .9
23
V
5 0 .9
24
Cr
5 2 .0
25
Mn
5 4 .9
26
Fe
5 5 .8
27
Co
5 8 .9
28
Ni
5 8 .7
29
Cu
6 3 .5
30
Zn
6 5 .4
13
Al
2 7 .0
31
Ga
6 9 .7
10
Ne
2 0 .2
18
Ar
3 9 .9
32
Ge
7 2 .6
33
As
7 4 .9
34
Se
7 9 .0
35
Br
7 9 .9
36
Kr
8 3 .8
37
Rb
8 5 .5
38
Sr
8 7 .6
39
Y
8 8 .9
40
Zr
9 1 .2
41
Nb
9 2 .9
42
Mo
9 5 .9
43
Tc
(9 8 )
44
Ru
101
45
Rh
103
46
Pd
106
47
Ag
108
48
Cd
112
49
In
115
50
Sn
119
51
Sb
122
52
Te
128
53
I
127
54
Xe
131
55
Cs
133
56
Ba
137
57
La
139
72
Hf
178
73
Ta
181
74
W
184
75
Re
186
76
Os
190
77
Ir
192
78
Pt
195
79
Au
197
80
Hg
201
81
Tl
204
82
Pb
207
83
Bi
209
87
Fr
(2 2 3 )
88
Ra
226
89
Ac
227
8B
85
86
84
At
Rn
Po
(2 0 9 ) ( 2 1 0 ) ( 2 2 2 )
104
105
106
107
108
110
109
Rf
Db
Sg
Bh
Hs
Ds
Mt
(2 6 1 ) (2 6 2 ) (2 6 3 ) (2 6 2 ) (2 6 5 ) (2 6 6 ) (2 8 1 )
Lant ha nid e s
58
Ce
140
59
Pr
141
Act inid e s
90
Th
232
91
Pa
231
60
Nd
144
92
U
238
61
62
63
64
65
66
67
68
69
70
71
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
(1 4 5 ) 1 5 0
152
157
159
162
165
167
169 173
175
93
94
95
96
97
98
99
100
101
102
103
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
( 2 3 7 ) ( 2 4 4 ) ( 2 4 3 ) ( 2 4 7 ) ( 2 4 7 ) ( 2 5 1 ) ( 2 5 2 ) ( 2 5 7 ) (2 5 8 ) ( 2 5 9 ) (2 6 0 )