Solutions to the Final Examination
202-001-50
Fall 2010
1. a) hydrochloric acid
mercury (I)
nitric acid
calcium fluoride
lead (IV) sulfate
barium chloride dihydrate
dinitrogen pentoxide
sodium perchlorate
b) CO32−
HBr (aq)
SF6
SnO
KMnO4
AgCℓ
CuS
NH4NO3
2.
Isotope
Symbol
Name
Sodium
23
11
Sulfur
Na
32
16
Chromium (III)
S
52
24
Cr
Phosphorus anion
31
15
P
3+
3−
Atomic
Number
Mass
Number
Number of
Neutrons
Number of
protons
Number of
electrons
11
23
12
11
11
16
32
16
16
16
24
52
28
24
21
15
31
16
15
18
3. a) 0.24 mL
0.025 g
25.205 cm
3
2.85 g  10 mg   10 mL  2.85 × 106 mg
b)


=
mL  g   L 
L
3
9
−297.4° F
( −183.0°C ) + 32 =
5
c)
−183.0°C + 273.15 =
90.2 K
d)
( 2.24 ×10 ) × ( 5.464 ×10 ) − 0.200 =
1.2 − 0.200 =
1.0
−5
e)
f)
4
1.0 × 102
V=
2.40 × 102 cm3 =
2.40 × 102 mL
( 5.00 cm ) × ( 4.00 cm ) × (12.0 cm ) =
 1.28 g 
2
2
m =
 × ( 2.40 × 10 mL ) =3.07 × 10 g =0.307 kg
 mL 
4.57×10–2
5.10×100
7.87×105
4. a) 4 Ga (s) + 3 O2 (g) → 2 Ga2O3 (s)
b) C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (ℓ)
c) 3 Mg (s) + Mn2O3 (s) → 3 MgO (s) + 2 Mn (s)
5. 6 CℓO2 (g) + 3 H2O (ℓ) → 5 HCℓO3 (aq) + HCℓ (ℓ)
Molar mass: CℓO2 = 67.45
H2O = 18.02
HCℓO3 = 84.46
142.0 g
38.0 g
= 2.105
=
mol nH O = 2.109 mol
67.45 g / mol
18.02 g / mol
=
nCO
2
2
a) Assume H2O is the Limiting Reactant.
 6 mol C O2 
nCO == 2.109 mol H 2 O 
 = 4.218 mol C O2 > 2.105 mol
 3 mol H 2 O 
The assumption is not valid. CℓO2 is the Limiting Reactant
2
 5 mol HC O3 
2.105
=
mol C O2 
 1.754 mol HC O3
6 mol C O2 

b)
 84.46 g 
=
mHCO 1.754
=
mol HC O3 
 148.2 g
 mol 
80.0 g
× 100%= 53.99%
c) % yield=
148.2 g
nHCO
3
3
6. a)
V1 V2
=
T2 T2
V2 =V1 ×
PV
=
RT
=
n
b)
T2
 100.0 + 273.15 
=1.88 L 
 =2.38 L
T1
 21.0 + 273.15 
(1.35 atm )( 2.50 L )
=
( 0.08206 L ⋅ atm ⋅ K ⋅ mol ) ( 303.2 K )
−1
−1
0.136 mol
 20.2 g 
=
=
m 0.136
mol 
 2.74 g
 mol 
c) C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (ℓ)
 3 mol O2 
0.500
=
mol C2 H 4 
 1.50 mol O2
 1 mol C2 H 4 
−1
−1
nRT (1.50 mol ) ( 0.08206 L ⋅ atm ⋅ K ⋅ mol ) ( 273.15 K )
=
= 33.6 L
V =
1.000 atm
P
nO
2
7. a) 8 Hg (ℓ) → S8 (s) → 8 HgS (s)
i) Mixture
HgS (s)
ii) Chemical change. Nature of the material has changes. As well the leading statement says
there was a reaction.
13.6 g
× 100.0 mL =
1360 g
mL
iii) m =
b) filtration, distillation, evaporation
c) A compound is a pure substance. A mixture is a physical combination of several pure
substances.
d) A homogenous mixture is uniform in appearance and composition. A heterogeneous mixture is
non-uniform in appearance and composition; there is a physical boundary between the
components of a heterogeneous mixture.
8. a)
Elements
Ti, Fe, Ag
Mg, Sr, Ba
Li, K, Rb
Ne, Kr, Xe
F, Br, I
b) i)
ii)
iii)
iv)
9.
Classification
Transition metals
Alkaline earth metals
Alkali metals
Noble gases
halogens
False. Neutrons have no charges
False. Electrons occupy the space outside the nucleus.
False. Most of the atomic mass resides in the nucleus.
False. Isotopes of an element may have different numbers of neutrons.
Acid-Base
Precipitation
Combustion
H3PO4 (aq) + 3 NaOH (aq) → Na3PO4 (aq) + 3 H2O (ℓ)
CaCℓ2 (aq) + Na2SO4 (aq) → 2 NaCℓ (aq) + CaSO4 (s)
2 C6H11OH (l) + 17 O2 (g) → 12 CO2 (g) + 12 H2O (ℓ)
i)
ii)
Fe2(SO4)3
P 4O 6
iii) (NH4)2C2O4
iv) SrCr2O7
Fe : +3
P: +3
N: –3
Sr: +2
S: +6
O: –2
H: +1
Cr: +6
O: –2
C: +3
O: –2
10. SiCℓ 4 (ℓ) + Mg (s) → MgCℓ2 (s) + Si (s)
SiCℓ 4 (ℓ) + 4 e– → Si (s) + 4 Cℓ– (aq)
Mg (s) + 2 Cℓ– (aq) → MgCℓ2 (s) + 2 e–
O: –2
reduction
oxidation
Balanced: SiCℓ 4 (ℓ) + 2 Mg (s) → Si (s) + 2 MgCℓ2 (s)
Oxidizing Agent: SiCℓ 4
Reducing Agent: Mg
Oxidized species: Mg
Reduced species: Si
11. Na2CO3 (aq) + 2 HCℓ (aq) → 2 NaCℓ (aq) + H2O (ℓ) + CO2 (g)
CIE: 2 Na+ (aq) + CO32− (aq) + 2 H+ (aq) + 2 Cℓ– (aq) → 2 Na+ (aq) + 2 Cℓ– (aq) + H2O (ℓ) + CO2 (g)
NIE: 2 H+ (aq) + CO32− (aq) → H2O (ℓ) + CO2 (g)
Spectator ions: Na+, Cℓ–
12. a)
Atom
C
H
N
O
Mass (g)
0.2990
0.05849
0.2318
0.1328
Moles (mol)
0.02489
0.05803
0.01655
0.008300
Mole Ratio
2.999
6.992
1.994
1.000
Simplest Ratio
3
7
2
1
Empirical Formula: C3H7N2O
b)
Molar Mass
= 1 Formula Mass: 3 (12) + 7 (1) + 2 (14) + 16 = 87
Formula Mass
Molecular Formula: C3H7N2O
13. a) 8 H2S (aq) + 8 Cℓ2 (aq) → 16 HCℓ (aq) + S8 (s)
 mol 
 2.40 × 10−5 g 
−5
n=
=
 3.52 × 10 mol
H S

 ( 50.0 L ) 
L
34.08
g




1.80 g
=
= 0.0254 mol
nC
70.91 g / mol
2
2
By inspection, the Limiting Reactant is H2S
5
=
nC 0.0254 mol − 3.52 × 10−=
mol 0.0254 mol
2
mC = 1.80 g
2
 1 mol S8   256.5 g 
mS =
3.52 × 10−5 mol H 2 S 
0.00113 g

=
 8 mol H 2 S   mol 
b)=
Actual Yield (18.65
=
g )( 0.400 ) 7.46 g
8
14. A: Exactly 50°C
B: Between 10°C and 50°C
C: Between 10°C and 50°C
15. a) m
=


0.220 M )(1.50 L ) 
(=

( 0.1 M )(1.5 L )
294.3 g
 mol 
97.1 g
b) V =
=
9 × 10−4 L =
0.9 mL
16 M
c) i) NaOH
ii) H2SO4
iii)
H2SO4 (aq) + 2 NaOH (aq) → Na2SO4 (aq) + 2 H2O (ℓ)
( 0.141 M )( 0.0142 L )  2 mol NaOH 
iv)
=
[ NaOH ] =

 0.0700 M
0.0572 L
 1 mol H 2 SO4 
16. a) i)
ii)
b) i)
ii)
The rate of the forward reaction is equal to the rate of the reverse reaction
KP =
PO2
PC4 2
N2 (g) + 3 H2 (g) → 2 NH3 (g)
2
PNH
3
KP =
PN2 PH32
[ NH ]=
( 0.157 )
= 6.02 × 10
[ N ][ H ] ( 0.921)( 0.763)
[ NH ]=
( 0.203)
= 6.02 × 10
[ N ][ H ] ( 0.399 )(1.197 )
[ NH ]=
(1.82 )
= 6.02 × 10
[ N ][ H ] ( 2.59 )( 2.77 )
2
No.1
=
KC
2
−2
3
3
2
3
2
2
iii)
No.2
=
KC
2
3
3
2
2
2
No.3
=
KC
3
2
−2
3
3
2
3
2
d) ii) Because the temperature of the reaction is constant.
17. a)
b)
d [H2 ]
d [ Br2 ]
1 d [ HBr ]
R=
−
=
+
=
+
dt
dt
2 dt
0.500 M − 0.455 M 0.045 M
Average
=
Rate
= = 0.0030 M / s
15 s
15 s
−2