Review Material for Exam #2
1.
a.
Calculate the molarity of a solution made with 184.6 mg sample of potassium
dichromate dissolved in enough water to give 500.0 mL of solution.
1 mole K 2 Cr2 O 7
1g
x
1000 mg
294.181 g K 2 Cr2 O7
= 1.255 x 10–3 M
1L
500.0 mL x
1000 mL
184.6 mg K 2 Cr2 O7 x
b.
What is the molarity of potassium ion?
K2Cr2O7(aq)
2 K+(aq) + Cr2O72–(aq)
1.255 x 103 mole K 2 Cr2 O7
2 mole K +
x
= 2.510 x 10–3 M K+
L
1 mole K 2 Cr2 O7
2.
Calculate the mass of sodium hydroxide needed to make 250.0 mL of a 0.4000 M
solution.
0.4000 mole NaOH
39.997 g NaOH
= 4.000 g NaOH
x 0.2500 L x
L
1 mole NaOH
3.
How would you prepare 1.0 L of 0.50 M sulfuric acid from concentrated (18 M) sulfuric
acid?
0.50 mole H 2SO 4
1L
1000 mL
x 1.0 L x
x
L
18 mole H 2SO 4
1L
= 28 mL concentrated H2SO4 diluted to 1.0 L
4.
Predict the solubility of each substance in water using solubility rules. Write the
ionization reaction when each soluble substance is dissolved in water.
1.
All Na+, K+, and NH4+ salts are soluble.
2.
All NO3–, C2H3O2–, ClO3–, and ClO4– salts are soluble.
3.
All Ag+, Pb2+ , and Hg22+ salts are insoluble.
4.
All Cl– , Br– , and I– salts are soluble.
5.
All CO32–, O2–, S2–, OH–, SO32–, CrO42–, Cr2O72–, and PO43– salts insoluble,
except CaS, SrS, BaS and Ba(OH)2.
6.
All SO42– salts are soluble except Ca2+, Sr2+, and Ba2+.
a.
b.
c.
Ba(NO3)2
ZnS
(NH4)2CO3
soluble (2)
Ba(NO3)2(aq)
insoluble (5)
soluble (1)
(NH4)2CO3(aq)
Ba2+(aq) + 2 NO3–(aq)
2 NH4+(aq) + CO32–(aq)
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
n.
o.
p.
q.
r.
s.
t.
u.
5.
Fe2O3
PbCl2
NaClO4
BaCrO4
FeBr2
Ag2SO4
Ca(OH)2
Hg2(ClO3)2
K3PO4
Mg3(PO4)2
NH4NO3
ZnCrO4
NiSO4
AlCl3
AgI
CaBr2
CuS
K2SO3
insoluble (5)
insoluble (3)
soluble (1)
insoluble (5)
soluble (4)
insoluble (3)
insoluble (5)
soluble (2)
soluble (1)
insoluble (5)
soluble (1)
insoluble (5)
soluble (6)
soluble (4)
insoluble (3)
soluble (4)
insoluble (5)
soluble (1)
Na+(aq) + ClO4–(aq)
NaClO4(aq)
Fe2+(aq) + 2 Br–(aq)
FeBr2(aq)
Hg22+(aq) + 2 ClO3–(aq)
3 K+(aq) + PO43–(aq)
Hg2(ClO3)2(aq)
K3PO4(aq)
NH4+(aq) + NO3–(aq)
NH4NO3(aq)
NiSO4(aq)
AlCl3(aq)
Ni2+(aq) + SO42–(aq)
Al3+(aq) + 3 Cl–(aq)
CaBr2(aq)
Ca2+(aq) + 2 Br–(aq)
K2SO3(aq)
2 K+(aq) + SO33–(aq)
Predict the products (include phase state), balance and write the net ionic equations for
each of the following. Classify each as precipitation, acid-base or oxidation-reduction.
Na3PO4(aq) +
3 AgNO3(aq)
3 Ag+(aq) + PO43–(aq)
K2SO4(aq) + BaCl2(aq)
Ba2+(aq) + SO42–(aq)
Hg2(NO3)2(aq)
+
+
AgClO3(aq) + KCl(aq)
Ag+(aq) + Cl–(aq)
3 KOH(aq)
+
BaSO4(s)
H3PO4(aq)
H3PO4(aq) + 3 OH–(aq)
precipitation
+
2 KCl(aq)
BaSO4(s)
precipitation
Hg2Cl2(s)
+ Ca(NO3)2(aq)
Hg2Cl2(s)
Na2SO4(aq)
Pb2+(aq) + SO42–(aq)
+ 3 NaNO3(aq)
Ag3PO4(s)
CaCl2(aq)
Hg22+(aq) + 2 Cl–(aq)
Pb(NO3)2(aq)
Ag3PO4(s)
precipitation
PbSO4(s)
+ 2 NaNO3(aq)
PbSO4(s)
AgCl(s) +
precipitation
KClO3(aq)
AgCl(s)
precipitation
K3PO4(aq)
+
3 H2O(ℓ)
PO43–(aq)
+
3 H2O(ℓ)
acid-base
ZnCl2(aq) + H2S(aq)
ZnS(s)
Zn2+(aq) + S2–(aq)
2 Na3PO4(aq)
+
+
2 HCl(aq)
ZnS(s)
3 CaCl2(aq)
precipitation
Ca3(PO4)2(s)
3 Ca2+(aq) + 2 PO43–(aq)
Ca3(PO4)2(s)
2 HBr(aq) + Ca(OH)2(aq)
CaBr2(aq) +
H+(aq) + OH–(aq)
2 Na(s) + S(s)
+ 6 NaCl(aq)
precipitation
2 H2O(ℓ)
H2O(ℓ)
acid-base
Na2S(s)
same
redox
2 C2H6(g) +
7 O2(g)
4 CO2(g)
+
6 H2O(g or ℓ)
same
redox
Mg(s) + Cu(NO3)2(aq)
Mg(s) + Cu2+(aq)
6.
Cu(s) + Mg(NO3)2(aq)
Mg2+(aq) + Cu(s)
redox
Determine the oxidation state of each element and identify which element is oxidized
and which is reduced. Identify the oxidizing and reducing agents.
a.
b.
c.
Fe2O3 + 3 CO
+3 –2
+2 –2
PbS + 4 H2O2
+2 –2
+1 –1
2 Fe + 3 CO2
0
+4 –2
PbSO4 + 4 H2O
+2 +6 –2 +1 –2
8 H2O + 3 P4 + 20 HNO3
+1 –2
0
+1 +5 –2
Fe reduced, C oxidized
Fe2O3 oxidizing agent
CO reducing agent
O reduced, S oxidized
H2O2 oxidizing agent
PbS reducing agent
12 H3PO4 + 20 NO
+1 +5 –2
+2 –2
N reduced, P oxidized
HNO3 oxidizing agent
P4 reducing agent
d.
2 MnSO4 + 5 PbO2 + 3H2SO4
+2 +6 –2
+4 –2 +1 +6 –2
2 HMnO4 + 5 PbSO4 + 2 H2O
+1 +7 –2
+2 +6 –2
+1 –2
Pb reduced, Mn oxidized
PbO2 oxidizing agent
MnSO4 reducing agent
e.
3 Cu + 2 HNO3 + 6 HCl
0
+1 +5 –2
+1 –1
3 CuCl2 + 2 NO + 4 H2O
+2 –1
+2 –2
+1 –2
N reduced, Cu oxidized
HNO3 oxidizing agent
Cu reducing agent
f.
2 Bi(OH)3 + 3 K2SnO2
+3 –2 +1
+1 +2 –2
3 K2SnO3 + 2 Bi + 3 H2O
+1 +4 –2
0
+1 –2
Bi reduced, Sn oxidized
Bi(OH)3 oxidizing agent
K2SnO2 reducing agent
7.
If 30.0 mL of 0.150 M aluminum chloride is added to 15.0 mL of 0.100 M silver(I)
nitrate, what mass of precipitate will be formed?
3 AgNO3(aq) + AlCl3(aq)
3 AgCl(s) + Al(NO3)3(aq)
0.150 mole AlCl3
3 mole AgCl 143.321 g AgCl
= 1.94 g AgCl
x 0.030 L x
x
L
1 mole AlCl3
1 mole AgCl
0.100 mole AgNO3
1 mole AgCl
143.321 g AgCl
= 0.215 g AgCl
x 0.015 L x
x
L
1 mole AgNO3
1 mole AgCl
limiting
8.
A 10.00 mL sample of barium hydroxide solution was titrated with 31.24 mL of
0.1250 M HCl. Calculate the molarity of the barium hydroxide solution.
2 HCl(aq) + Ba(OH)2(aq)
BaCl2(aq) + 2 H2O(ℓ)
0.1250 mole HCl
1 mole Ba(OH) 2
x 0.03124 L x
L
2 mole HCl
= 0.1953 Ba(OH)2
0.01000 L
9.
Calcium in blood is determined by precipitating calcium oxalate, CaC2O4, followed by
dissolving in acid and titrating the oxalic acid (H2C2O4) with KMnO4:
5 H2C2O4(aq) + 2 MnO4–(aq) + 6 H+(aq)
10 CO2(g) + 2 Mn2+(aq) + 8 H2O(ℓ)
What is the concentration of Ca2+ (mg/dL) in a 10.0 mL sample of blood if 10.54 mL of
9.88 x 10–4 M KMnO4 solution is needed for the titration.
9.88 x 104 mole KMnO 4
1 mole MnO 4 
5 mole H 2C 2O 4
x
x 0.01054 L x
L
1 mole KMnO 4
2 mole MnO 4 
0.01000 L
x
10.
1 mole Ca 2+
40.078 g Ca 2+ 1000 mg
1L
x
x
x
= 10.4 mg/dL
2+
1 mole H 2 C2 O 4
1 mole Ca
1g
10 dL
Acetylene is a gas used as a fuel for some welding torches. If 0.52 L of acetylene has a
pressure of 1824 torr, what is the pressure (in atms) if the volume is decreased to 0.39 L?
Boyles Law: P1V1 = P2V2;
1 atm 

1824 torr x

  0.52 L 
P1V1
760 torr 

P2 

 3.2 atm
V2
0.39 L
11.
A sample of carbon dioxide has a volume of 19.4 L at 10.8 °C. What is the volume of the
same sample of carbon dioxide at 14.6 °C.
Charles Law:
12.
VT
V1
V
 2 ; V2  1 2 
T1
T1
T2
19.4 L  (14.6 + 273.15) K 
(10.8 + 273.15) K
 19.7 L
A balloon at 23 °C contains 0.32 moles of helium gas at 2432 torr. The volume of the
helium is 2.43 L. If an additional 0.14 mole of He is injected into the balloon while
holding the temperature and pressure constant, what is the volume of the balloon?
 2.43 L  0.46 mole  
Vn
V1
V
 2 ; V2  1 2 
3.5 L
n1
0.32 mole
n1
n2
A sample of Ar has a pressure of 0.63 atm at 26.4 °C and a volume of 0.79 L. If the
temperature is lowered by 5.2 °C and the pressure is increased to 0.96 atm, what is
the new volume?
Avogadro’s Law:
13.
P1V1
PV
PVT
 2 2 ; V2  1 1 2 
P2T1
T1
T2
 0.63 atm  0.79 L  (26.4  5.2) + 273.15
 0.96 atm  26.4+ 273.15
 0.53 L
14.
Magnesium metal reacts with hydrochloric acid to produce magnesium chloride and
hydrogen gas. What mass of Mg reacted with excess hydrochloric acid if 255 mL of
hydrogen gas is produced at 756.4 torr and 25.0 oC?
1 atm
 0.9640 atm
P  756.4 torr  23.76 torr  x
760 torr
1L
V 255 mL x
 0.255 L
1000 mL
n ?
L  atm
R 0.08206
mole  K
T 25.0 oC + 273.15 = 298.15K
PV = nRT
 0.9640 atm  0.255 L 
PV
n

= 0.01005 mole H2
L  atm 
RT

 0.08206
  298.15 K 
mole  K 

Mg(s) + 2 HCl(aq)
0.01005 mole H 2 x
15.
MgCl2(aq) + H2(g)
1 mole Mg
24.305 g Mg
= 0.244 g Mg
x
1 mole H 2
1 mole Mg
Styrene oxide is a fairly simple aromatic organic compound that has a pleasant odor and
is often used in the perfume industry. If 2.07 g of the compound is vaporized completely
into a closed 1.04 L flask at 435 ºC, the pressure in the flask is found to be 735 torr.
Calculate the formula weight of styrene oxide.
1 atm
P 735 torr x
 0.967 atm
760 torr
V 1.04 L
g 2.07 g
L  atm
R 0.08206
mole  K
T 435 oC + 273.15 = 708.15 K
FW ?
PV 
FW 
gRT
FW
gRT

PV
L  atm 
  708.15 K 
mole  K 

= 120 g/mole (3 sig. figs.)
 0.967 atm 1.04 L 
 2.07 g   0.08206
16.
a.
A sample of dichloroethane, a dry cleaning solvent, is vaporized into a 266.4 mL
flask at 99.8 oC. Some of the sample leaves the flask through a pinhole
through the stopper until the pressure in the flask is the same as the outside
pressure. A barometer shows the pressure to be 745.3 torr. When the flask is
cooled, the mass of dichloroethane in the flask is measured to be 0.8447 g.
Calculate the formula weight of dichloroethane.
1 atm
P 745.3 torr x
 0.9807 atm
760 torr
1L
V 266.4 mL x
 0.2664 L
1000 mL
g 0.8447 g
L  atm
R 0.08206
mole  K
T 99.8 oC + 273.15 = 372.95 K
FW ?
PV 
gRT
FW
gRT
FW 

PV
b.
L  atm 
  372.95 K 
mole  K 

= 98.95 g/mol
 0.9807 atm  0.2664 L 
 0.8447 g   0.08206
Is the formula of dichloroethane CH2Cl2, C2H4Cl2, or C4H8Cl2?
84.933 g/mole, 98.960 g/mole, 127.014 g/mole
17.
Diborane, B2H6, is a highly explosive compound formed by the reaction
3 NaBH4(s) + 4 BF3(g)
2 B2H6(g) + 3 NaBF4(s)
What mass of NaBH4 is required to give 1.00 L of diborane at 0.0 oC and 1.00 atm?
P
V
n ?
R
T
1.00 atm
1.00 L
L  atm
mole  K
0.0 oC + 273.15 = 273.15 K
0.08206
PV = nRT n 
1.00 atm 1.00 L 
PV

= 0.0446 mole B2H6
L  atm 
RT

 0.08206
  273.15 K 
mole  K 

0.0446 mole B2 H 6 x
3 mole NaBH 4
37.833 g NaBH 4
= 2.53 g NaBH4
x
2 mole B2 H 6
1 mole NaBH 4
18.
For the reaction, as written:
S8(s) + 8 O2(g)
a.
8 SO2(g)
How much heat is evolved when 25 moles of sulfur is burned in excess oxygen?
25 mole S8 x
b.
2368 kJ
= –5.9 x 104 kJ
1 mole S8
How much heat is evolved when 275 grams of sulfur is burned in excess oxygen?
275 g S8 x
c.
1 mole S8
2368 kJ
= –2.54 x 103 kJ
x
256.52 g S8 1 mole S8
How much heat is evolved when 150.0 grams of sulfur dioxide is produced?
150.0 g SO 2 x
19.
H = –2368 kJ
1 mole S8
1 mole SO 2
2368 kJ
= –693.1 kJ
x
x
64.063 g SO 2
8 mole SO 2 1 mole S8
In a calorimetry experiment, 0.1277 g of Mg ribbon was added to 200.0 mL 0.500 M
HCl at 24.12 oC. The water temperature increased to 27.10 oC. Calculate H for this
reaction, as performed, and H per mole of HCl.
H   CT   (m H2O x s H2O )T   (200.0 g x 4.184
J
)(27.10 o C  24.12 o C)
g C
o
H = –2493.66 J
Mg(s) + 2 HCl(aq)
0.1277 g Mg x
MgCl2(s) + H2(g)
1 mole Mg
2 mole HCl
x
= 0.0105 mole HCl needed
24.305 g Mg
1 mole Mg
0.500 mole HCl
x 0.2000 L = 0.100 mole HCl available; Mg limiting
L
H 
2493.66 J
0.0105 mole HCl
 –237,491 J/mole = –237 kJ/mole HCl
20.
A 2.50 g sample of sucrose (C12H22O11) was burned in excess oxygen in a calorimeter
which contained 2.19 kg of water. The temperature of the water increased from 20.50 oC
to 25.01 oC. Determine the molar heat of combustion of sucrose.
C12H22O11(s)
+
12 O2(g)
12 CO2(g) + 11 H2O(g)
H  CT  (m H2O x s H2O )T   (2190 g x 4.184
J
)(25.01 o C  20.50 o C)
g C
o
H = –41,324.9 J
2.50 g C12 H 22 O11 x
H 
1 mole C12 H 22 O11
= 7.30 x 10–3 mole C12H22O11
342.297 g C12 H 22 O11
41,324.9 J
7.30 x 10 3 mole C12 H 22 O11
 –5,658,162.51 J/mole
= –5.66 x 103 kJ/mole C12H22O11
21.
The burning of 5.08 g benzene (C6H6) releases enough heat to raise the temperature of
5.00 kg of water from 10.1 oC to 19.6 oC. Calculate the molar heat of combustion of
benzene.
2 C6H6(ℓ) + 15 O2(g)
12 CO2(g)
+
H  CT  (m H2O x s H2O )T   (5000 g x 4.184
6 H2O(g)
J
)(19.6 o C  10.1 o C)
g C
o
H = –198,740 J
5.08 g C6 H 6 x
H 
1 mole C6 H 6
= 0.065 mole C6H6
78.114 g C6 H 6
198,740 J
0.065 mole C6 H 6
 –3,055,979.6 J/mole = –3.1 x 103 kJ/mole C6H6
22.
Given the following equations and H values,
H3BO3(aq)
HBO2(aq) + H2O(ℓ)
H2B4O7(s) + H2O(ℓ)
H2B4O7(s)
4 HBO2(aq)
2 B2O3(s) + H2O(ℓ)
H
–0.02 kJ
–11.3 kJ
+17.5 kJ
calculate Horxn for the following reaction: 2 H3BO3(aq)
2 H3BO3(aq)
2 HBO2(aq) + 2 H2O(ℓ)
2(–0.02 kJ)
2 HBO2(aq)
1/2 H2B4O7(s) + 1/2 H2O(ℓ)
–½(–11.3 kJ)
B2O3(s) + 1/2 H2O(ℓ)
½(+17.5 kJ)
+14.4 kJ
1/2 H2B4O7(s)
23.
B2O3(s) + 3 H2O(ℓ)
The following reaction is one that occurs in a blast furnace when iron is extracted from
its ores:
Fe2O3(s) + 3 CO(g)
2 Fe(s) + 3 CO2(g)
find Horxn for this reaction given the following information:
3 Fe2O3(s) + CO(g)
FeO(s) + CO(g)
Fe3O4(s) + CO(g)
Fe2O3(s) + 1/3 CO(g)
2 FeO(s) + 2 CO(g)
2/3 Fe3O4(s) + 2/3 CO(g)
2 Fe3O4(s) + CO2(g)
Fe(s) + CO2(g)
3 FeO(s) + CO2(g)
H
–46.4 kJ
9.0 kJ
–41.0 kJ
2/3 Fe3O4(s) + 1/3 CO2(g)
2 Fe(s) + 2 CO2(g)
1/3(–46.4 kJ)
2(9.0 kJ)
6/3 FeO(s) + 2/3 CO2(g) 2/3(–41.0 kJ)
–24.8 kJ
24.
Calculate Horxn for the following reaction:
C2H4(g) + H2(g)
C2H6(g)
from the following information:
C2H4(g) + 3 O2(g)
2 CO2(g) + 2 H2O(ℓ)
C2H6(g) + 7/2 O2(g)
H2(g) + 1/2 O2(g)
2 CO2(g) + 3 H2O(ℓ)
H2O(ℓ)
C2H4(g) + 3 O2(g)
H2(g) + 1/2 O2(g)
25.
–1550 kJ
–286 kJ
2 CO2(g) + 2 H2O(ℓ)
2 CO2(g) + 3 H2O(ℓ)
H
–1401 kJ
–1401 kJ
C2H6(g) + 7/2 O2(g) –(–1550 kJ)
H2O(ℓ)
–286 kJ
–137 kJ
Give the formation reactions for the following compounds: Al2O3, NH3, N2O5, C2H3Cl.
2 Al(s) + 3/2 O2(g)
Al2O3(s)
1/2 N2(g) + 3/2 H2(g)
N2(g) + 5/2 O2(g)
NH3(g)
N2O5(g)
2 C(s) + 3/2 H2(g) + 1/2 Cl2(g)
C2H3Cl(ℓ)