Chemistry Reference
In-Depth Information
11.4
The two solutions have the same initial concentration. When
they are combined, the final concentration is also the same.
(The two initial solutions and the final solution would all taste
equally sweet. Try it.) As an analogy, if a car travels 50 mph
during the first half of a trip and 50 mph during the second
half of the trip, its overall speed is 50 mph. Alternatively, we
may calculate the final concentration:
MnO
2
2 KClO
3
(s)
----
¡
10.125 (a)
(b)
Since the catalyst the and the KCl are
all solids, the loss in mass is equal to the mass of the
gas
produced—oxygen. The oxygen therefore has a mass of
2 KCl(s)
3 O
2
(g)
(MnO
2
),
KClO
3
,
1.221 g
0.998 g
0.223 g
(c)
The mass of that
reacted
can be calculated from
the mass of oxygen produced:
KClO
3
a
1.22 mol
1 L
0.0250 L
b
0.0305 mol
a
2 mol KClO
3
3 mol O
2
a
122.5 g KClO
3
1 mol KClO
3
a
1 mol O
2
32.0 g O
2
0.223 g O
2
b
b
b
a
1.22 mol
1 L
0.0350 L
b
0.0427 mol
0.569
1
g KClO
3
(d)
The mass of KCl produced can also be calculated from
the mass of oxygen produced:
0.0732 mol
0.0600 L
1.22 M
a
74.5 g KCl
1 mol KCl
a
1 mol O
2
32.0 g O
2
a
2 mol KCl
3 mol O
2
0.365 mol
0.4000 L
0.912
5
M
0.223 g O
2
b
b
b
11.5
(a)
(b)
Because 365 mmol is 0.365 mol, this problem is the same
as (a).
(c)
Because 400.0 mL is 0.4000 L, this problem is the same
as (a).
0.346
1
g KCl
Alternatively, the mass of KCl produced can be calcu-
lated from the mass of
KClO
3
decomposed minus the
mass of oxygen produced:
365 mmol
400.0 mL
0.912
5
M
0.365 mol
0.4000 L
0.569
1
g
0.223 g
0.346
1
g
(d)
(e)
At least 0.569 g of
KClO
3
must have been present initially.
a
1 mol NH
3
17.0 g NH
3
11.6
The concentration of the portion is 0.693 M, the same as the
concentration of the original sample. (Tea would taste as
sweet if sipped from a 100.0 mL portion or from 3.13 mL
poured directly from that 100.0 mL portion.)
10.127
20.2 g NH
3
b
1.18
8
mol NH
3
a
1 mol HNO
3
63.0 g HNO
3
41.6 g HNO
3
b
0.660
3
mol HNO
3
NH
3
(aq)
HNO
3
(aq)
£
NH
4
NO
3
(aq)
11.7
(a)
2 dozen brides, 2 dozen grooms
(b)
2.0 M NH
4
,
2.0 M NO
3
Initial:
1.18
8
mol
0.660
3
mol
0.000 mol
Change:
Final:
0.660
3
mol
0.660
3
mol
0.660
3
mol
1.0 M Na
1.0 M Cl
11.8
(a)
and
(b)
and
(c)
and
(d)
and
(e)
and
(f)
and
Note that the subscript after the O in the formulas in parts
(d)-(f ) represents the number of oxygen atoms per nitrate ion,
not the number of nitrate ions per formula unit.
(g)
0.52
8
mol
0.000 mol
0.660
3
mol
1.0 M Mg
2
2.0 M Cl
a
14.0 g N
1 mol N
1.0 M Cr
3
3.0 M Cl
2 mol N
1 mol NH
4
NO
3
a
b
b
18.5 g N
0.660
3
mol NH
4
NO
3
1.0 M Li
1.0 M NO
3
1.0 M Co
2
2.0 M NO
3
10.131
This reaction proceeds because of the formation of the
(largely covalent) acetic acid.
Ba(C
2
H
3
O
2
)
2
(aq)
2 HCl(aq)
1.0 M Al
3
3.0 M NO
3
£
BaCl
2
(aq)
2HC
2
H
3
O
2
(aq)
Initial:
0.150 mol
0.500 mol
0.000 mol
0.000 mol
2.0 M Al
3
3.0 M SO
4
2
and
Change:
Final:
0.150 mol
0.300 mol
0.150 mol
0.300 mol
2.0 M NH
4
1.0 M SO
4
2
(h)
and
0.000 mol
0.200 mol
0.150 mol
0.300 mol
10.132
$233.00
CT
4 Ch £ Set
11.9 (a)
The final volume is about 5.5 L.
(b)
The final volume is 3.3 L. Note the small difference in the
wording of the parts of this problem, which makes a large
difference in the answer.
11.10 (a)
A chemical reaction occurs.
(b)
The number of moles of nitrate ion must be calculated
from the numbers of moles in the two solutions.
(c)
The number of moles of chloride ion must be calculated.
(d)
A chemical reaction occurs.
(e)
Initial:
5000.00
31
98
0
Change:
21
84
21
Final:
107.00 10 14 21
The manager can buy only 21 sets; money (as usual) is the
limiting quantity. Here, three “reactants” have to be consid-
ered to see which is limiting.
4893.00
11 Molarity
11.1
Each solution merely dilutes the concentrations of the
ions in the other.
(a)
3.000 M (The small sample is the same concentration as
the original solution because it has not been changed.)
(b)
2.000 mL
(c)
(f)
The number of moles of hydroxide ion in the final solu-
tion must be calculated.
6.000 mmol (one-thousandth of the 6.000 mol in the sample)
11.11 (a)
To determine the concentration of the
NaHCO
3
solution.
(d)
6.000 mmol/2.000 mL
3.000 mmol/mL
3.000 M
2.542 M
(b)
To determine the number of moles of HA, and from that
the molar mass of HA. 57.81 g/mol
1.70 mol
1.00 L
1.70 mol
2.00 L
1.70 M
0.850 M
11.2
(a)
(b)
1.70 mol
5.00 L
1.70 mol
0.500 L
3.40 M
H
CO
3
2
(c)
0.340 M
(d)
11.12 (a)
The ions react with the ions, forming water
and carbon dioxide, two covalent compounds, and leav-
ing fewer ions in solution.
(b)
There is no interaction between the ions in these
compounds.
11.3
(a)
Less than 40.0 mL of water is used.
(b)
Exactly 40.0 mL of water is used, and more than 40.0 mL
of solution is produced.
