CHM 134 General Chemistry I
Exam 2, Fall 2007 – Dr. Steel
Name
SOLUTIONS
1. (10 points) Circle T or F to indicate whether each statement is true or false.
F
The oxidation number of sulfur in sulfur trioxide is +3.
T
A mixture of aqueous solutions of cobalt (II) chloride and sodium phosphate
will conduct an electric current.
T
The reaction between sodium iodide and lead (II) nitrate forms a solid.
F
According to the balanced equation
2 Mg(s) + O2(g)→ 2 MgO(s)
combining 4 grams of magnesium with 2 grams of oxygen should produce 6
grams of magnesium oxide.
T
The oxidation number of carbon in carbon disulfide is positive four.
F
The oxidizing agent in a reaction serves the purpose of donating electrons to
the substance being treated.
T
200. mL of 0.100-M aluminum chloride, AlCl3, contains 0.0600 moles of
chlorine ions.
T
If tin and oxygen react to produce tin (IV) oxide, the reducing agent is the tin.
F
An element can not be both reduced and oxidized in the same reaction.
F
When you are given two reaction quantities, the limiting reactant is always the
reactant of which you have fewer moles.
2. (5 points) Indicate whether each species is soluble or insoluble in water.
SOLUBLE
FeCl3
INSOLUBLE
ZnS
INSOLUBLE
BaCO3
SOLUBLE
INSOLUBLE
MgO
INSOLUBLE
Fe(OH)3
SOLUBLE
H3PO4
SOLUBLE
Na2SO4
SOLUBLE
Al(NO3)3
SOLUBLE
LiF
NH4Br
3. (4 points) Indicate the oxidation number of each element in these species.
S2O32-
C2H6O
-2 C
+1 H -2 O
+2 S
KClO4
-2 O
+1 K +7 Cl
-2 O
4. (12 points) Balance each equation by providing the appropriate coefficients. If you
leave a blank before a compound it will be interpreted as a coefficient of 1.
3 NH4NO3 +
2 Al →
MnO2 + 4 HCl →
3 N2 +
MnCl2 +
2 Na + 2 H2O →
6
4
S2Cl2 +
16 NH3 →
C3H5N3O9 →
12
6 H2O +
Cl2 + 2 H2O
2 NaOH +
12 NH4Cl +
CO2 + 10
Al2O3
H2O +
H2
S8 +
S4N4
O2 + 6
N2
5. (6 points) Determine the products and write the balanced molecular chemical
equation for the reaction between aqueous solutions of lead (II) sulfite and
hydrochloric acid. Be sure to include phase notations.
PbSO3(aq) + 2 HCl(aq) → PbCl2(s) + H2O(l) + SO2(g)
6. (8 points) What volume of 0.164-M hydrochloric acid is required to neutralize
250. mL of solution that contains 22.6 grams of potassium hydroxide?
1 mol KOH
= 0.4028 moles KOH
56.11 g KOH
moles HCl = M HCl VHCl = moles KOH
moles KOH = 22.6 g KOH ×
VHCl =
moles KOH 0.4028 moles
=
= 2.46 L HCl
M HCl
0.164 mol
L
7. (10 points) Nitric oxide is made from the oxidation of ammonia. What mass of nitric
oxide can be produced when 28.5 g of ammonia are combined with 34.9 g of oxygen?
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
28.5 g NH 3 ×
1 mol NH 3
4 mol NO 30.01 g NO
×
×
= 50.2 g NO
17.03 g NH 3 4 mol NH 3 1 mol NO
34.9 g O 2 ×
1 mol O 2
4 mol NO 30.01 g NO
×
×
= 26.2 g NO
32.00 g O 2 5 mol O 2
1 mol NO
Therefore, only 26.2 g of NO can be produced.
8. (8 points) What mass of the excess reactant remains after the reaction in question 7 is
completed?
34.9 g O 2 ×
4 mol NH 3 17.03 g NH 3
1 mol O 2
×
×
= 14.9 g NH 3 used
32.00 g O 2
5 mol O 2
1 mol NH 3
28.5 g − 14.9 g = 13.6 g NH 3 leftover
9. (6 points) Indicate whether each reaction is a precipitation, acid-base, gas forming, or
redox reaction. (You ay use the space at the right to determine the products of the
reactions, but do not need to provide them.)
REDOX
2 Sn(s) + O2(g) →
GAS FORMING
Na2CO3(aq) + HCH3CO2(aq) →
PRECIPITATION
NH4I(aq) + Pb(NO3)2(aq) →
ACID-BASE
NaOH(aq) + HBr(aq) →
PRECIPITATION
NaCl(aq) + CuNO3(aq) →
GAS FORMING
Li2S(aq) + 2 HClO3(aq) →
10. (10 points) An aqueous solution of chloric acid (HClO3) has a pH of 3.15. What mass
of HClO3 is present in 1.00 L of this solution?
pH = −log[H + ]
[H + ] = 10 − pH = 10 −3.150 = 7.08 × 10 − 4 M
[H + ] = [HClO 3 ]
moles HClO 3 = M × V = (7.08 × 10 − 4 M )(1.00 L) = 7.08 × 10 − 4 moles HClO 3
7.08 × 10 − 4 moles HClO 3 l ×
84.46 g HClO 3
= 0.0598 g HClO 3
1 mol HClO 3
11. (5 points) Balance this redox reaction and indicate which species is the oxidizing
agent and which is the reducing agent.
C2H2(l) +
2 Cl2(g) →
oxidizing agent
Cl2
C2H2Cl4(l)
C in C2H2
reducing agent
12. (6 points) What mass of NaBrO4 is required to prepare a 250. mL sample of 0.358 M
solution?
moles NaBrO 4 = M × V = (0.358 M )(0.250 L ) = 0.0895 mol NaBrO 4
0.0895 mol NaBrO 4 ×
166.89 g NaBrO 4
= 14.9 g NaBrO 4
1 mol NaBrO 4
13. (10 points) The reaction below is known to have a typical percent yield of 74.8%. If
you recover 8.17 grams of silver phosphate, what volume of 0.135-M silver fluoride
was used in the reaction?
3 AgF(aq) + Na3PO4(aq) → Ag3PO4(s) + 3 NaF(aq)
8.17 g Ag 3 PO 4 = 0.748(Theoretical Yield )
8.17 g Ag 3 PO 4
= 10.9 g Ag 3 PO 4
0.748
1 mol Ag 3 PO 4
3 mol AgF
10.9 g Ag 3 PO 4 ×
×
= 0.0781 mol AgF used
418.67 g Ag 3 PO 4 1 mol Ag 3 PO 4
Theoretical Yield =
M=
mol
mol 0.0781 mol AgF
⇒V=
=
= 0.579 L = 579 mL
V
M
0.135 mol
L