Exam # ___
Name
(last)
(First-Name)
Signature
Exam 2 Fall 2007
Read all questions before you start. Show all work and explain your answers to receive full credit. Report all numerical answers to the proper
number of significant figures. By signing your signature above, you agree that you are adhering to the academic honesty policy. If you cannot
adhere to this policy you should drop this course. Keep your eyes on your own paper at all time
Multiple choice. (4 pts each) Show work for partial credit.
1
What conversion factor is required to convert mass of specie A to volume of specie A in a stoichiometry problem?
a) molarity
b) molar mass
c) Avogadro’s Number
d) density
e) mole fraction
2
3
Which quantity is based on 100% conversion of reactant to products in a chemical reaction?
a) theoretical yield
b) percent yield
c) actual yield
d) titration yield
e) no right answer
Ammonia is formed in a direct reaction of nitrogen and hydrogen:
N2(g) + H2(g) → NH3(g) The starting mixture is represented in the big diagram to the
right in which the dark circles represent N and the white circles represent H.
Which of the following represents the product mixture?
a)
b)
c)
d)
4
Which compound will not support electricity when added to pure distilled water?
a) sodium chloride
b) cadmium chloride
c) lead(II) chloride
d) hydrochloric acid
e) no right answer
5
What is conserve (does not change) in the dilution equation, M1 • V1 = M2 • V2 ?
a) mass solvent
b) molarity solute
c) volume solution
d) moles solute
e) no right answer
6
Which is always found in the buret during a titration experiment?
a) analyte
b) indicator
c) titrant
d) base
7
What is the identity of this precipitate when barium nitrate is combined with potassium sulfate?
a) Ba(NO3)2
b) BaSO4
c) K2SO4
d) KNO3
e) solute
e) N2SO4
8
For real gases, what parameter must be adjusted in the ideal gas law equation because of the incorrect assumption in the KMT, which
states that "particles do not attract or repel each other"?
a) volume
b) pressure
c) temperature
d) density
e) moles
9
In a gas mixture of H2, Ne and Ar with a total pressure of 7.20 atm, what is the mole fraction of H2 if the respective partial
pressure of Ne and Ar are 0.80 and 1.00 atm ?
a) 0.75
b) 0.33
c) 0.25
d) 1.5
e) none
10
How many moles of Fe are needed to produce 10.0 mol of H2?
a) 10.0 mol
b) 13.3 mol
c) 15.0 mol
4H2O(g) + 3Fe(s) → Fe3O4(s) + 4H2(g)
d) 30.0 mol
e) 7.50 mol
11
(20 pts) Balance the following equations with proper phases. Convert chemical names to chemical formulas first for iv and v.
i) __2_ ZnS(s) + __3_ O2(g) → __2_ ZnO (s) + __2_ SO2 (g)
__2_ ZnO (s) + __2_ SO2 (g) → __2_ ZnS(s) + __3_ O2(g)
ii) __1__PbS (aq) + __4__H2O2 (aq) → __1__ PbSO4 (aq) + __4__H2O (l)
__1__ PbSO4 (aq) + __4__H2O (l) → __1__PbS (aq) + __4__H2O2 (aq)
iii) __1__ ClO4- + __8__ I- + __8__H+ →
__4__ I2 + __1__ Cl- + __4_ H2O
__4__ I2 + __1__ Cl- + __4_ H2O → __1__ ClO4- + __8__ I- + __8__H+
iv) Solid ammonium nitrate breaks down to dinitrogen monoxide gas and dihydrogen monoxide liquid.
.
__1_ NH4NO3 (s) → __1_ N2O (g) + __2_ H2O (l).
v) Solid sodium oxide and liquid water forms aqueous sodium hydroxide
.
__1_ Na2O (s) + __1_ H2O (l). → __2_ NaOH (aq)
12
(8 pts) When taking a SCUBA class, the instructor should always points out the first rule of SCUBA:
[A diver must always breathe continuously under water and especially upon the ascent]
This is an important rule because the lungs can only withstand an over-pressure of approximately 2 psi. The pressure in seawater
increases 0.5 psi per foot depth and thus in just a four foot ascent, a diver would experience an increase of 2 psi in his lungs if he
fails to exhale. 14.7 psi = 1 atm = 760 torrs
Consider a scenario in which a non-trained diver did not know this rule. If she were to breathe 5 pints of air at 99 ft (4 atm, 40°F)
and ascend to the surface (1 atm, 65°F) without exhaling, theoretically, what would be the volume of gas in his lungs (in pints)?
Combine gas law, 40°F = 277.4K, 65°F = 291.3K
Her lungs would be expanding at
P1V1 n1RT1
=
P2V2 n2RT2
0.16 pints / feet.
V2 =
"
P1V1 T1
PV T
=
" V2 = 1 1 2
P2V2 T2
P2T1
If her lungs were rigid,
then the pressure in her lungs would in crease
0.0303 atm / feet or 0.445 psi / ft
P1V1T2 P1 (4 atm) # V1 (5 pints) # T2 (291.2K)
=
P2T1
P2 (1atm) T1 (277.4K)
Her lungs will burst at 4.5 ft into her assend or
V2 = 21 pints
13
!
at a level of 94.5 ft depth.
(7 pts) In a titration, 43.5 ml of 0.500 M Ca(OH)2 is added dropwise to 17.4 ml HClO4 solution until the equivalent point is reached.
!
Write the balance equation and then calculate the molar concentration of the HClO4 solution?
Ca(OH)2
(aq)
+ 2HClO4
MHClO4 = 0.500
"
2H2O (l) + Ca(ClO4 )2
(aq)
mol
2 mol HClO4
1
Ca(OH)2 # 43.5 ml #
#
L
1 mol Ca(OH)2 17.4 ml
MHClO4 = 2.50 M HClO4
!
(aq)
14
(18 pts each) Check mark if you agree___ or disagree___ with the statements below and write a complete justification for your
answer. You will not receive any credit without justifying your answer.
i) agree__ or disagree_X_
All gases have equal densities since according to the KMT, gases are treated as point particles moving in random motion.
No, D = MP /RT, so the density is a function of the molar mass of a gas
ii) agree_ X_ or disagree__
For a gas at constant temperature, decreasing the volume of its container increases its pressure.
Yes, According to Boyle’s Law there is an inverse relationship between pressure and volume of a gas, the smaller the volume the
higher the pressure at constant temperature for a fix amount of gas
iii) agree_ X_ or disagree__
In the reaction of Fe + O2 → Fe2O3, iron is the reducing agent.
Yes, In the reaction, iron is oxidized to +3 state, the oxidized species give up its electron and therefore is responsible for reducing
its partner, thus it is a reducing agent.
iv) agree__ or disagree_ X_
In a chemical equation such as those in question 11, the subscript (s) stands for the soluble phase.
No, (s) as a phase means solid, not soluble.
v) agree__ or disagree_ X_
Since HI is a molecule with covalent bonds, it will not ionize and therefore is a non-electrolyte.
No, Eventhough HI is a covalent compount, it is a strong acid and delivers its H+ ion in solution, the increase in ion defines that this
chemical is a strong electrolyt.
vi) agree__ or disagree_ X_
The end point in a titration is where the mass of the acid is equal to the mass of the base.
No, in a titration, the end point generally means that the moles of the acid is equal to the moles of the base..
15
(15pts) The reaction of SO2 with Cl2 to give dichlorine monoxide occurs as follow:
SO2(g) + 2 Cl2(g) → SOCl2(g) + Cl2O(g)
All of the compounds involved in the reaction are gases at 300 K.
i) Calculate the Urms of each chemical and place them in order of increasing average speed
Lowest speed to highest speed
ii) Draw a Maxwell-Boltzmann distribution curve for these gases with the relative Urms shown.
U rms =
MWSO2
3RT
7482.6
, at 300 K Urms =
M
M
= 64 g/mol , MWCl2 = 71 g/mol
U rms (Cl2 ) =
MWSOCl2 = 119 g/mol , MWCl2 O = 87 g/mol
U rms (SO2 ) =
!
U rms (SOCl2 ) =
7482.6
m
= 1.17•105 = 342
s
-3 kg
64•10
mol
U rms (Cl2O) =
!
7482.6
m
= 1.05•105 = 325
s
-3 kg
71•10
mol
7482.6
m
= 0.629•105 = 251
s
-3 kg
119•10
mol
7482.6
m
= 0.860•105 = 293
s
-3 kg
87•10
mol
16
4pts
(24 pts) A solution of silver nitrate is to be prepared by carefully weighing 8.50 grams of silver nitrate, placing it in a 500-ml
volumetric flask and filling to the mark with water.
i) What is the concentration of this stock solution? Label this Cs. (MW: silver nitrate = 169.88 g/mol)
Concentration of Stock, C s
C s = 8.50 g AgNO3 "
3pts
!
mol
1
"
= 0.100 M
169.88 g
0.500L
ii) A second solution is to be prepared from this stock solution in which 10.00 ml of the stock solution is transferred to a 100-ml
volumetric flask and then filled to the mark with water. What is the concentration of this second solution? Label this C1.
C s " Vs = C1 " V1 # C1 =
C1 =
9pts
!
AgNO 3
C s " Vs
0.100M "10.0mL
=
V1
100. mL
0.100M "10.0mL
= 0.0100M
100. mL
iii) An aliquot of 50.00ml of the C1 solution is added to 0.100 grams of potassium iodide. Write the chemical reaction that occur in
terms of the a) molecular equation, b) the complete ionic equation, and c) the net ionic equation.
a) Molecular equation: AgNO3 (aq) + KI (aq) → KNO3(aq) + AgI(s)
b) Complete ionic equation: Ag+ (aq) + NO3- (aq) + K+ (aq) + I- (aq) →
b) Net ionic equation: Ag+ (aq) + I- (aq)
2pts
AgI(s)
iv) Identify the precipitate and spectator ions.
Spectator ions: K+ (aq)
Precipitate: AgI(s)
6pts
→
+ NO3- (aq)
v) Calculate the mass of the product formed from the reaction.
Moles AgNO3 = 0.0100M " 0.050 L = 5.00 "10 -4 mol AgNO3
Moles KI = 0.100g KI "
mol
= 6.02 •10 -4 mol KI
166. g
Limiting is the AgNO3
Amount of AgI formed is from 5.00"10 -4 mol AgNO3 (limiting)
Mass AgI = 5.00 "10 -4 mol AgI "
!
K+ (aq) + I- (aq) + AgI(s)
234.8 g
= 0.118 g AgI
1 mol
17
(18pts) Gaseous iodine pentafluoride, IF5, can be prepared by the reaction of solid iodine and gaseous fluorine:
I2 (s) + 5F2 (g) → 2IF5 (g). A 5.00 L flask is charged with 10.0 g I2 and 10.0 g F2, and the reaction proceeds until one of the
reagents is completely consumed. After the reaction is complete the temperature in the flask is 125°C.
i) What is the partial pressure of IF5 in the flask ?
ii) What is the mole fraction of IF5 in the flask ?
I2 (MW) = 253.8
g
g
, F2 (MW) = 38.0
,
mol
mol
Mole I2 = 10.0g I2 "
mol
= 0.0394 mol I2
253.8 g
Mole F2 = 10.0g F2 "
mol
= 0.263 mol F2
38.0 g
4 pts
Limiting is I2 , Calculate mol IF5 produced mol IF5 = 0.0394 mol I2 "
4 pts
2mol IF5
= 0.0788 mol IF5
1mol I2
Amount F2 Reacted
mol F2 = 0.0394 mol I2 "
4 pts
5mol F2
= 0.197 mol F2
1mol I2
Excess F2 Remaining
4 pts
Excess F2 = 0.263 mol F2 - 0.197 mol F2 = 0.066 mol
Partial Pressure IF5 , pIF5
.0788 mol IF5 " 0.08206
pIF5 =
5.0 L
L " atm
" 398 K
mol"K
= 0.515 atm
Mole fraction IF5
# IF5 =
18
mol IF5
0.0788
=
= 0.544
mol IF5 + mol F2
0.07885 + 0.066
(10 Bonus pts) Calculate the greatest quantity of matter.
i) 0.255 L!
of Au (ρ = 19.3 g/ml)
moles of Au 0.255 L = 255 ml
ii) 560. L Ar at 1atm, 0°C
moles of Ar at STP
iii) 5.00L of 5.00 M of Acetic Acid
moles Arsenic
moles acetic acid, HAc
mol
19.3g 1mol
" 5.00 L = 25.0 mol HAc
255 mL "
"
= 25.0 mol Au 560 L " 1mol Ar = 25.0 mol Ar 5.00
L
mL 197 g
22.4 L
!
!
iv) 1.51•1026 atoms of As
!
!
1 mol As atoms
=
6.02"1023 atoms
moles Arsenic = 251 mol As atoms
1.51"1026 atoms "