CHAPTER FOUR
TYPES OF CHEMICAL REACTIONS AND SOLUTION
STOICHIOMETRY
Questions
9.
"Slightly soluble" refers to substances that dissolve only to a small extent. A slightly soluble salt
may still dissociate completely to ions and, hence, be a strong electrolyte. An example of such a
substance is Mg(OH)2' It is a strong electrolyte, but not very soluble. A weak electrolyte is a
substance that doesn't dissociate completely to produce ions. A weak electrolyte may be very
soluble in water, or it may not be very soluble. Acetic acid is an example of a weak electrolyte
that is very soluble in water.
10.
Measure the electrical conductivity of a solution and compare it to the conductivity of a solution
of equal concentration of a strong electrolyte.
Exercises
Aqueous solutions: Strong and Weak Electrolytes
11.
12.
Na+(aq) + Br"(aq)
a.
NaBr(s)
c.
AI(N03Ms)
e.
HI(g)
g.
KMnOis)
i.
NH4C2HP2(S) -+ NH/(aq) + C 2H30 2'(aq)
a.
c.
-+
-+
b.
AI 3+(aq) + 3 N0 3'(aq)
MgClis)
-+
Mi+(aq) + 2 CI'(aq)
d. (NH4)2S0is)
W(aq) + I'(aq)
-+
2 NH/(aq) + SO/'(aq)
Fe2+(aq) + SO/'(aq)
f.
FeS04(s)
h.
HCIOis) -+ H+(aq) + CI04'(aq)
HCI(g) -+ H+(aq) + CI'(aq)
b.
HN03(l)
Ca(OHMs) -+ Ca2+(aq) + 2 OH'(aq)
d. KOH(s) -+ K+(aq) + OH'(aq)
-+
-+
K+(aq) + Mn04'(aq)
13.
CaClb) -+ Ca2+(aq) + 2 CI"(aq)
14.
MgSOis)
---+
Mg 2+(aq) + SO/,(aq); NH4N03(s)
60
-+
-+
-+
H+(aq) + N03'(aq)
NH/(aq) + N0 3'(aq)
CHAPTER 4
61
SOLUTION STOICHIOMETRY
Solution Concentration: Molarity
15.
a.
5.623 g NaHC03 x
1mol NaHCO
3 = 6.693
84.01 gNaHC03
10-2 mol NaHC03
X
2
M = 6.693 x 10- mol x 1000 mL = 0.2677 MNaHC0
3
250.0mL
L
°
1 mol~Cr
2 7 = 6.275
294.20 g ~Cr207
b. 0.1846 g K2CrP7 x
X
10-4 mol K2Cr20 7
4
M= 6.275 x 10- mol = 1.255 x 10-3 MK 2Crp7
500.0 x 10-3 L
c. 0.1025 g Cu x 1 mol Cu = 1.613
X
10-3 mol Cu = 1.613
X
10-3 mol Cu2+
63.55 g Cu
3
M= 1.613 x 10- mol Cu 2+ x 1000 mL = 8.065 x 10-3 MCu2+
200.0mL
16.
a.
L
16.45 g NaCI x 1 mol NaCI = 0.2815 MNaCI
1.000 L
58.44 g NaCI
b. 853.5 mg KI0 3 x
1
1 mol KIO
g
x
3 = 3.988
1000 mg 214.0 g KI03
X
10-3 mol KI03
3
3.988 x 10- mol x 1000 mL = 1.595 x 10-2MKI0
3
250.0mL
L
c. 0.4508 g Fe x
I mol Fe = 8.072 x 10-3 mol Fe = 8.072 x 10-3 mol Fe3+
55.85 g Fe
3
2
8.072 x 10- mol Fe + x 1000 mL = 1.614 x 10-2 MFe3+
500.0mL
L
17.
a.
CaClis)~Ca2+(aq)+2CI-(aq);
MCa 2+ =0.15M; Met - =2(0.15)=0.30M
b. AI(N03Ms) ~ AI3+(aq) + 3 N03'(aq); MAI ]+ = 0.26 M; MNO - = 3(0.26) = 0.78 M
]
c. K2Cr20 7(s) ~ 2 K+(aq) + Cr20/"(aq); MK += 2(0.25) = 0.50 M; MCr20 ;- = 0.25 M
d. AliS04Ms) ~ 2 AI3+(aq) + 3 SO/'(aq)
MAll+=
2.0xl0-3moIAliS04)3
L
x
2molAl 3+
=4.0
X
_
10 3 M
mol AllS04)3
2.0 x 10-3 mol AliS04)3
3 mol sot
M S0 2- =
x
== 6.0
4
L
mol AliS04)3
X
10-3 M
CHAPTER 4
62
18.
0.100 mol Ca(N03)2
a.
MCa(NO) =
2.5 mol N~SO4
=2.0M
1.25 L
b.
M
c.
5.00 g NH4CI x
=
N~S04
MNH CI
=
1 molNH4CI
53.49 g NH4 CI
0.0935 mol NH4CI
0.5000L
4
d.
1.00 g K 3P04 x
M
=
K 3P04
1.00 M
=
0.100 L
32
19.
SOLUTION STOICHIOMETRY
=
=
0.0935 mol NH4CI
0.187M
1 molK PO
3__4 = 4.71 x 10-3 mol K3P04
212.27 g
3
4.71 x 10- mol = 0.0188 M
0.2500 L
mol solute = volume (L) x molarity ( m:l) ; AICI 3(s)
mol CI- = 0.1000 L x
0.30 mol AICI 3
x
L
MgC1b)
~
~
=
9.0
3
X
10-2 mol CI­
molAICI3
Mg2+(aq) + 2 CI-(aq)
mol CI- = 0.0500 L x
NaCI(s)
3 mol CI ­
~ AI +(aq) + 3 CI"(aq)
0.60 mol MgCI 2
L
x
2 mol CI ­
mol MgCI 2
= 6.0 x 10"2 mol CI­
Na+(aq) + CI-(aq)
mol CI- = 0.2000 L x 0.40 mol NaCI x 1 mol CI - = 8.0
L
mol NaCI
X
10-2 mol CI­
100.0 mL of 0.30 M AICI 3 contains the largest moles of CI" ions.
CHAPTER 4
20.
SOLUTION STOICHIOMETRY
63
NaOH(s) -.. Na+(aq) + OH"(aq), 2 total mol of ions (1 mol Na+ and 1 mol Cn per mol NaOH.
0.1000 L x 0.100 mol NaOH x 2 mol ions = 2.0 x 10'2 mol ions
L
mol NaOH
BaClz(s) -.. Ba2+(aq) + 2 CI'(aq), 3 total mol of ions per mol BaCI 2 •
0.0500 Lx 0.200 mol x 3 mol ions = 3.0 x 10-2 mol ions
L
mol BaCl z
0.0750 L x
0.150 mol Na3P04
4 mol ions
x
L
mol Na3P04
.
= 4.50 x 10'2 mol IOns
75.0 mL of 0.150 MNa 3P0 4 contains the largest number of ions.
21.
Molar mass ofNaHC03 = 22.99 + 1.008 + 12.01 + 3(16.00) = 84.01 g/mol
Volume
= 0.350 g NaHC03 x
1 mol NaHC0 3
84.01 g NaHC0 3
x
1L
0.100 mol NaHC0 3
= 0.0417 L = 41.7 mL
41.7 mL of 0.100 MNaHC0 3 contains 0.350 g NaHC0 3 •
22.
Molar mass of NaOH = 22.99 + 16.00 + 1.008 = 40.00 g/mol
Mass NaOH = 0.2500 L x 00400 mol NaOH x 40.00 g NaOH = 4.00 g NaOH
L
mol NaOH
23.
a.
2.00 L x 0.250 mol NaOH x 40.00 g NaOH = 20.0 g NaOH
L
mol
Place 20.0 g NaOH in a 2 L volumetric flask; add water to dissolve the NaOH and fill to the
mark with water, mixing several times along the way.
b. 2.00 L x
0.250 mol NaOH
1 L stock
= 0.500 L
x
L
1.00 mol NaOH
Add 500. mL of 1.00 MNaOH stock solution to a 2 L volumetric flask; fill to the
mark with water, mixing several times along the way.
c. 2.00 L x
0.100 mol ~Cr04
L
x
194.20 g ~Cr04
mol KzCr0 4
As in a, instead using 38.8 g K 2Cr0 4 •
= 38.8 g K 2Cr04
CHAPTER 4
64
d. 2.00 L x
0.100 mol ~crO 4
x
L
SOLUTION STOIClllOMETRY
1 L stock
= 0.114 L
1.75 mol ~Cr04
As in b, instead using 114 mL of the 1.75 MK zCr0 4 stock solution.
24.
a.
1.00 L solution x
0.50 mol H2S04
= 0.50 mol HZS0 4
L
1L
= 2.8 x 10- z Leone. HZS0 4 or 28 mL
18 mol H2S04
0.50 mol HZS0 4 x
Dilute 28 mL of concentrated HZS0 4 to a total volume of 1.00 L with water.
b. We will need 0.50 mol HCI.
1L
= 4.2
12 mol HCl
0.50 mol HCI x
X
lO- z L = 42 mL
Dilute 42 mL of concentrated HCl to a final volume of 1.00 L.
c.
We need 0.50 mol NiCl z.
0.50 mol NiCl z x
1 mol NiCI 2·6Hp
mol NiCl 2
x
237.69 gNiCl 2·6H20
mol NiCl 2·6H20
=
118.8 g NiCl z.6HzO :::: 120 g
Dissolve 120 g NiCl z.6HzO in water, and add water until the total volume of the solution is
1.00 L.
d. 1.00 L x
0.50 mol HN03
=
L
0.50 mol HN03
1L
= 0.031 L = 31 mL
16 mol HN03
0.50 mol HN03 x
Dissolve 31 mL of concentrated reagent in water. Dilute to a total volume of 1.00 L.
e.
We need 0.50 mol NazC03'
0.50 mol NaZC03 x
105.99 gN~C03
mol
=
53 g NazC03
Dissolve 53 g NaZC03 in water, dilute to 1.00 L.
25.
1 mol
= 8.17
132.15 g
10.8 g (NH4)zS04 x
X
lO' z mol (NH 4)zS04
2
Molarity = 8.17 x 10- mol x 1000 mL = 0.817 M(NH )z S0 4
4
100.0 mL
L
Moles of (NH4)2S04 in final solution
10.00
x
10-3 L
x
0.817 mol = 8.17
L
x
10-3 mol
CHAPTER 4
65
SOLUTION STOICHIOMETRY
. =
· a ffima I so IutlOn
MoIanty
(NH4)zS04(S)
---+
3
8.17 x 10- mol
(10.00 + 50.00) mL
1000
x -mL
L
2 NH/(aq) + SOt(aq); M NH + = 2(0.136)
4
26.
mol NazCO J = 0.0700 L
3.0 mol Na2C03
x
= 0.136 M(NH4 )zS0 4
= 0.272 M; Msoz.
= 0.136 M
4
= 0.21 mol NazCOJ
L
NazCOJ(s) ~ 2 Na+(aq) + CO/"(aq); mol Na+ = 2(0.21) = 0.42 mol
mol NaHCO J = 0.0300 L
NaHCOJ(s)
= 0.030 mol NaHC03
L
Na+(aq) + HCOJ'(aq); mol Na+ = 0.030 mol
= total mol Na + 0.42 mol + 0.030 mol
M
Na +
27.
---+
1.0 mol NaHC0 3
x
total volume
·
Stock so Iut IOn
100.0
0.45 mol = 4.5 MNa+
0.1000 L
0.0700 L + 0.0300 L
3
2.00 x 10- 5 g steroid
= 10.0 mg = 10.0 x 10- g = ------'''----­
500.0 mL
500.0 mL
mL
5
X
10-6 L stock x 1000 mL x 2.00 x 10- g steroid
L
mL
=
2.00 x 10-6 g steroid
This is diluted to a final volume of 100.0 mL.
2.00
28.
6
10- g steroid
100.0 mL
x
x
1000 mL
L
- - - x
1 mol steroid
336.43 g steroid
594
=.
x
10-8 M st erOl'd
Stock solution:
1.584 g Mn z+ x 1 mol Mn 2+
54.94 g Mn 2+
= 2.883
2
10-2 mol Mnz+; 2.883 x 10- mol Mn 2+
1.000 L
X
Solution A contains:
50.00 mL x
2
1L
x 2.883 X 10- mol
1000 mL
L
=
1.442 x lO- J mol Mn 2+
3
Molarity = 1.442 x 10- mol x 1000 mL = 1.442 x 10-J M
1000.0 mL
L
Solution B contains:
3
I_L_ x 1.442 x 10- mol
10.0 mL x _ _
1000 mL
L
5
Molarity = 1.442 x 10- mol
0.2500 L
=
5.768
x
=
1.442 x 10-5 mol Mn z+
10-5 M
= 2.883
X
lO- z M
CHAPTER 4
66
SOLUTION STOICHIOMETRY
Solution C contains:
10.00
5
X
10-3 Lx 5.768 x 10- mol = 5.768 x 10-7 mol Mn 2+
L
7
Molarity = 5.768 x 10- mol = 1.154 x 10-6 M
0.5000 L
Precipitation Reactions
29.
In all these reactions, soluble ionic compounds are mixed together. To predict the precipitate,
switch the anions and cations in the two reactant compounds to predict possible products, then
use the solubility rules in Table 4.1 to predict if any of these possible products are insoluble (are
the precipitate).
a.
Possible products = BaS04 and NaCI; precipitate = BaSOis)
b. Possible products = PbCl2 and KN0 3; precipitate = PbClb)
c. Possible products = Ag3P04 and NaN03; precipitate = Ag3POb)
30.
d.
Possible products = NaN03 and Fe(OH)3; precipitate = Fe(OHMs)
a.
Possible products = FeCI 2 and K 2S04; Both salts are soluble so no precipitate forms.
b. Possible products = AI(OH)3 and Ba(N03)2; precipitate = AI(OHMs)
c. Possible products = CaS04 and NaCI; precipitate =CaSOis)
d.
31.
Possible products = KN0 3 and NiS; precipitate = NiS(s)
For the following answers, the balanced molecular equation is first, followed by the complete
ionic equation, then the net ionic equation.
a.
BaCI2(aq) + N~S04(aq)
---+
BaSOb) + 2 NaCI(aq)
Ba2+(aq) + 2 CI-(aq) + 2 Na+(aq) + SOt(aq)
Ba2+(aq) + SOt(aq)
b.
---+
---+
PbCI 2(s) + 2 KN0 3(aq)
Pb2+(aq) + 2 N0 3-(aq) + 2 K+(aq) + 2 CI-(aq)
---+
BaSOb) + 2 Na+(aq) + 2 CI-(aq)
BaSOis)
Pb(N03Maq) + 2 KCI(aq)
Pb2+(aq) + 2 CI-(aq)
---+
PbClb)
---+
PbClb) + 2 K+(aq) + 2 N0 3-(aq)
CHAPTER 4
d.
32.
SOLUTION STOICHIOMETRY
67
3 NaOH(aq) + Fe(N03Maq) ---+ Fe(OHMs) + 3 NaN03(aq)
a. No reaction occurs since all possible products are soluble salts.
b.
2 AI(N03Maq) + 3 Ba(OHMaq) ---+ 2 AI(OHMs) + 3 Ba(N03Maq)
AI 3+(aq) + 3 OH'(aq) ---+ AI(OHMs)
c.
CaCI2(aq) + N~S04(aq) ---+ CaSOb) + 2 NaCI(aq)
Ca2+(aq) + 2 CI'(aq) + 2 Na+(aq) + SOt(aq) ---+ CaSOb) + 2 Na+(aq) + 2 CI'(aq)
Ca2+(aq) + SOt(aq) ---+ CaSOb)
d.
K2S(aq) + Ni(N03Maq) ---+ 2 KN03(aq) + NiS(s)
2 K+(aq) + S2'(aq) + NF+(aq) + 2 N0 3' (aq) ---+2 K+(aq) + 2 N0 3'(aq) + NiS(s)
33.
a.
Silver iodide is insoluble. AgN03(aq) + KI(aq) ---+ AgI(s) + KN0 3(aq)
Ag+(aq) + I'(aq) ---+ AgI(s)
b.
Copper(II) sulfide is insoluble. CuS04(aq) + N~S(aq) ---+ CuS(s) + N~S04(aq)
c.
CoCI2(aq) + 2 NaOH(aq) ---+ Co(OHMs) + 2 NaCI(aq)
C02+(aq) + 2 OH(aq) ---+ Co(OHMs)
34.
d.
The potential products are Ni(N03)2 and KCI. Both are soluble in water. Thus, no reaction
occurs.
a.
AgCI is insoluble. Ag+(aq) + CI"(aq) ---+ AgCI(s)
b. FeS is insoluble. Fe2\aq) + S2'(aq) ---+ FeS(s)
c.
No reaction
d. Hg 2(N03)2 is made up of Hg/+ and N03' ions. Hg2C12, mercury(I) chloride or mercurous
chloride, is insoluble. Hg/+(aq) + 2 Cl'(aq) ---+ Hg 2CI2(s)
CHAPTER 4
68
35.
a.
SOLUTION STOICHIOMETRY
(NH4)2S0iaq) + Ba(N03)laq) ---. 2 NH4N0 3(aq) + BaSOis)
Ba2+(aq) + SO/(aq) ---. BaSOis)
b. Pb(N03)laq) + 2 NaCI(aq) ---. PbCI 2(s) + 2 NaN03(aq)
Pb 2\aq) + 2 CI-(aq)
-4
PbCI 2(s)
c. Potassium phosphate and sodium nitrate are both soluble in water. No reaction occurs.
d. No reaction occurs since all possible products are soluble.
e. CuCI 2(aq) + 2 NaOH(aq)
Cu2\aq) + 2 OR(aq)
36.
a.
-4
-4
Cu(OH)ls) + 2 NaCI(aq)
Cu(OH)ls)
CoCI 3(aq) + 3 NaOH(aq)
-4
Co(OH)ls) + 3 NaCI(aq)
C03\aq) + 3 OH-(aq) ---. Co(OH)ls)
b. 2 AgN0 3(aq) + (NH4)2C03(aq)
2 Ag\aq) + CO/"(aq)
-4
c. CuSOiaq) + HgCI 2(aq)
-4
Ag2C03(S) + 2 NH4N0 3(aq)
Ag 2C03(S)
-4
CuCI 2(aq) + HgS04(s)
Hg 2+(aq) + SO/(aq) ---. HgSOis)
d. No reaction occurs since all possible products (SrI 2 and KN0 3) are soluble.
37.
Three possibilities are:
Addition ofK2S04 solution to give a white ppt. ofPbS04. Addition ofNaCI solution to give
a white ppt. ofPbCl2. Addition ofK2Cr04 solution to give a bright yellow ppt. ofPbCr0 4.
38.
Since no precipitates formed upon addition ofNaCI or Na 2S04, we can conclude that Hg/+ and
Ba2+are not present since Hg2C1 2 and BaS04 are insoluble salts. Since a precipitate formed with
NaOH, the solution must contain Mn 2+which forms Mn(OH)ls).
39.
The reaction is: AgN0 3(aq) + NaCI(aq)
-4
AgCI(s) + NaNOiaq)
50.0 x 10-3 L AgN0 x 0.0500 mol AgN0 3 x I mol NaCI x 58.44 g NaCI
3
mol NaCI
1 mol AgN0 3
L AgN0 3
40.
= 0.146 g NaCI
2 Na3POiaq) + 3 Pb(N03)laq) ---. Pb3(P04)lS) + 6 NaN0 3(aq)
0.1500Lx
0.250 mol Pb(N03)2
L
x
2 mol Na3 P0 4
3 mol Pb(N03)2
x
1 L Na3P0 4
0.100 mol Na 3P0 4
= 0.250 L
CHAPTER 4
41.
SOLUTION STOICHIOMETRY
Al(N03)laq) + 3 KOH(aq)
0.0500 L x
~
69
Al(OH)ls) + 3 KN0 3(aq)
0.200 mol Al(NO )
33 = 0.0100 mol Al(N03)3
L
0.2000 L x 0.100 mol KOH = 0.0200 mol KOH
L
From the balanced equation, 3 mol ofKOH are required to react with 1 mol of Al(N03)3(3:1 mol
ratio). The actual KOH to Al(N03)3 mol ratio present is 0.020010.0100 = 2 (2: 1). Since the
actual mol ratio present is less than the required mol ratio, then KOH is the limiting reagent.
0.0200 mol KOH x
42.
1 mol Al(OH)3
78.00 g Al(OH)3
x
3 mol KOH
=
mol Al(OH)3
The balanced equation is: 3 BaCliaq) + Fe z(S04)laq)
100.0 mL BaClz x
IL
x
0.100 mol BaCI
1000 mL
~
2 = 1.00
X
L
0.520 g Al(OH)3
3 BaSOb) + 2 FeCI3(aq)
10-z mol BaClz
The required mol BaClz to mol Fez(S04)3 ratio from the balanced reaction is 3: I. The actual mol
ratio is 0.0 I 0010.0 I00 = I (1: I). This is well below the required mol ratio so BaClz is the
limiting reagent.
0.0100 mol BaCl z x
43.
a.
3 mol BaS04
3 mol BaCl 2
x
233.4 g BaS04
mol BaS04
=
2.33 g BaS04
The balanced equation is: 2 KOH(aq) + Mg(N03)z(aq)
~
Mg(OH)z(s) + 2 KN0 3(aq)
b. The precipitate is magnesium hydroxide.
c. 0.1000 L KOH x .0.200 mol KOH = 2.00 x 10-z mol KOH
LKOH
0.1000 L Mg(N03)Z x
0.200 mol Mg(NO )
3 2 = 2.00
LMg(N°3)2
X
lO- z mol Mg(N03)Z
From the balanced equation, the required mol KOH to mol Mg(N03)Z ratio is 2: I. The actual
mol ratio present is I: I. Not enough KOH is available to react with all of the Mg(N03)Z
present, so KOH is the limiting reagent.
0.0200 mol KOH x _I_m_o_lM_g_(O_H_)_2 x 58.33 g Mg(OH)2 = 0.583 g Mg(OH)z
2 mol KOH
mol Mg(OH)2
CHAPTER 4
70
d.
SOLUTION STOICrnOMETRY
The net ionic equation for this reaction is: Mg2+(aq) + 2 OR(aq)
---+
Mg(OHMs)
Since KOH was the limiting reagent, then all of the OR was used up in the reaction. So,
M OH ­ = 0 M. Note that K+ is a spectator ion, so it is still present in solution after
precipitation was complete. Also present will be the excess Mg z+ and N03' (the other
spectator ion).
mol Mg z+ initially
= 0.0200 mol Mg(N03)Z
mol Mg z+ reacted = 0.0200 mol KOH
x
1 mol Mg 2+
mol Mg(N03)2
x
1 mol Mg(N03)2
2 mol KOH
M
= 0.0200 mol Mg z+
1 mol Mg 2+
x
= 0.0100 mol Mg z+
mol Mg(N03)2
2
2
2+ = mol excess Mg + = 0.0200 - 0.0100 mol Mg + = 5.00
Mg
total volume
0.1000 L + 0.1000 L
x
lO' z MMg z+
The spectator ions are K+ and N0 3·. The moles of each present are:
mol K+
= 0.0200 mol KOH
x
1 mol K + = 0.0200 mol K+
molKOH
mol N0 3' = 0.0200 mol Mg(N03)Z
2 molN0 3­
x
mol Mg(N03)2
= 0.0400 mol N0 3'
The concentrations are:
= 0.0200 mol K + = 0.100 MK+' M
MK+
44.
'
0.2000 L
2 AgNOiaq) + CaClz(aq)
mol AgN0 3 = 0.1000 L
mol CaCl z = 0.1000 L
x
x
---+
NO
_= 0.0400 mol N0 3­
3
0.2000 L
= 0.200 MN0 3'
2 AgCl(s) + Ca(N03Maq)
0.20 mol AgN0 3
L
0.15 mol CaCI
2
L
=
0.020 mol AgN0 3
= 0.0 IS mol CaClz
The required mol AgN0 3 to mol CaCl z ratio is 2: 1 (from the balanced equation). The actual mol
ratio present is 0.020/0.015 = 1.3 (1.3:1). Therefore, AgN0 3 is the limiting reagent.
mass AgCl = 0.020 mol AgN0 3
x
I mol AgCl
I mol AgN0 3
x
143.4 g AgCl
mol AgCI
= 2.9 g AgCl
The net ionic equation is: Ag\aq) + CI'(aq) ---+ AgCI(s). The ions remaining in solution are the
unreacted Cl- ions and the spectator ions, N0 3- and Caz+(all Ag+is used up in forming AgCl).
The mol of each ion present initially (before reaction) can be easily determined from the mol of
each reactant. 0.020 mol AgN0 3 dissolves to form 0.020 mol Ag+ and 0.020 mol N0 3-. 0.015
mol CaClz dissolves to form 0.015 mol Caz+ and 2(0.015) = 0.030 mol Cl-.
CHAPTER 4
71
SOLUTION STOICHIOMETRY
mol unreacted C1- = 0.030 mol CI- initially - 0.020 mol C1- reacted = 0.010 mol C1- unreacted
= 0.010 mol CI- =
M
CI-
0.010 mol CI= 0.050 MCI-
0.1000 L + 0.1000 L
total volume
The molarity of the spectator ions are:
M
-=
NO]
0.020 mol N0 3-
= 0.10 MN0 3-;
0.2000 L
Mc
0015
2.
=.
a
I Ca 2+
mo
0.2000 L
= 0.075
MCa 2+
Acid-Base Reactions
45.
All the bases in this problem are ionic compounds containing OH-. The acids are either strong or
weak electrolytes. The best way to determine if an acid is a strong or weak electrolyte is to
memorize all the strong electrolytes (strong acids). Any other acid you encounter that is not a
strong acid will be a weak electrolyte (a weak acid) and should be kept together in a balanced
equation. The strong acids to recognize are HCI, HBr, HI, RN0 3, HCI0 4 and H2S04 • For the
answers below, the order of the equations are molecular, complete ionic and net ionic.
a. 2 HCIOlaq) + Mg(OH2)(s)
~
2 H20(l) + Mg(CI0 4Maq)
2 H+(aq) + 2 CI04-(aq) + Mg(OHMs)
~
2 H+(aq) + Mg(OHMs)
b. HCN(aq) + NaOH(aq)
HCN(aq) + OH"(aq)
~
c. HCI(aq) + NaOH(aq)
~
2 Hp(l) + Mg 2+(aq)
H 20(l) + NaCN(aq)
~
H20(l) + CN-(aq)
~
H 20(l) + NaCI(aq)
H+(aq) + Cl"(aq) + Na+(aq) + OH"(aq)
H+(aq) + OH"(aq)
46.
a.
~
2 H 20(I) + Mg2+(aq) + 2 CI04-(aq)
~
H 20(l) + Na+(aq) + CI-(aq)
H20(l)
3 HN0 3(aq) + A1(OHMs)
~
3 H20(l) + AI(N03Maq)
3 H+(aq) + 3 N0 3-(aq) + AI(OHMs)
3 H+(aq) + AI(OHMs)
~
~
3 Hp(l) + AI3+(aq) + 3 N0 3-(aq)
3 Hp(l) + AP+(aq)
CHAPTER 4
72
c. Ca(OHMaq) + 2 HCl(aq)
---+
2 HzO(I) + CaClz(aq)
Caz+(aq) + 2 OR(aq) + 2 H+(aq) + 2 Cl"(aq)
47.
---+
2 HzO(1) + Ca2+(aq) + 2 Cl"(aq)
HzO(I) + KN0 3(aq)
K+(aq) + OH"(aq) + H+(aq) + N0 3-(aq)
OR(aq) + H+(aq)
---+
---+
Hz0(l) + K+(aq) + N03"(aq)
HzO(I)
b. Ba(OHMaq) + 2 HCl(aq)
---+
2 Hz0(l) + BaClz(aq)
Baz+(aq) + 2 OR(aq) + 2 H+(aq) + 2 CI"(aq)
2 OH"(aq) + 2 H+(aq)
---+
---+
a.
AgOH(s) + HBr(aq)
2 HzO(l) + Baz+(aq) + 2 CI"(aq)
---+
HzO(l)
---+
3 HzO(1) + Fe3+(aq) + 3 CI0 4'(aq)
3 HzO(l) + Fe3+(aq)
AgBr(s) + Hz0(l)
--+
AgOH(s) + H+(aq) + Br-(aq)
---+
AgBr(s) + HzO(1)
AgOH(s) + H+(aq) + Br-(aq)
---+
AgBr(s) + Hz0(l)
b. Sr(OHMaq) + 2 HI(aq)
---+
2 HzO(1) + Srlz(aq) ,
Sr+(aq) + 2 OR(aq) + 2 H\aq) + 2 I"(aq)
Fe(OHMs) + 3 H+(aq) + 3 N0 3"(aq)
Fe(OHMs) + 3 H+(aq)
---+
3 HzO(I) + Fe(CI04 Maq)
3 H+(aq) + 3 Cl04 -(aq) + Fe(OHMs)
3 H+(aq) + Fe(OHMs)
---+
2 HzO(I) or OH"(aq) + H+(aq)
c. 3 HCI0 4(aq) + Fe(OHMs)
49.
---+
All the acids in this problem are strong electrolytes. The acids to recognize as strong electrolytes
are HCI, HBr, HI, HN0 3 , HCI0 4 and HZS0 4 •
a. KOH(aq) + HN0 3(aq)
48.
SOLUTION STOICHIOMETRY
---+
---+
---+
2 Hz0(l) + Sr+(aq) + 2 t(aq)
3 Hz0(l) + Fe3+(aq) + 3 N0 3"(aq)
3 HzO(1) + Fe 3+(aq)
Ifwe begin with 50.00 mL of 0.200 MNaOH, then:
50.00 x 10-3 L x 0.200 mol
L
= 1.00 x lO-z mol NaOH is to be neutralized.
CHAPTER 4
SOLUTION STOICHIOMETRY
73
a. NaOH(aq) + HCI(aq) --+ NaCI(aq) + H20(l)
1.00
X
IL
10.2 mol NaOH x I mol HCI x
= 0.100 L or 100. mL
mol NaOH 0.100 mol
b. RN03(aq) + NaOH(aq)
H20(l) + NaNOJCaq)
--+
1 mol RNO
1L
3 x
= 6.67 x 10.2 L or 66.7 mL
1.00 x 10-2 mol NaOH x
mol NaOH
0.150 mol RN03
c.
HC 2H30 2(aq) + NaOH(aq)
1.00
50.
X
--+
Hp(I) + NaC 2HP2(aq)
1 molHC H O I L
10-2 mol NaOH x
= 5.00
2 3 2x
mol NaOH
0.200 mol HC 2H30 2
X
10-2 L or 50.0 mL
We begin with 25.00 mL of 0.200 MHCI or 25.00 x 10.3 Lx 0.200 mol/L = 5.00 x 10-3 mol HCI.
a.
HCI(aq) + NaOH(aq)
5.00
X
--+
H20(l) + NaCI(aq)
10.3 mol HCI x I mol NaOH x
IL
= 5.00
mol HCI
0.100 mol NaOH
X
10.2 L or 50.0 mL
b. 2 HCI(aq) + Ba(OH)iaq) --+ 2 H20(l) + BaCI 2(aq)
5.00
c.
X
1 mol Ba(OH)
1L
2x
= 5.00
2 mol HCI
0.0500 mol Ba(OH)2
10-3 mol HCI x
HCI(aq) + KOH(aq)
--+
X
10-2 L = 50.0 mL
Hp(I) + KCI(aq)
IL
= 2.00 x 10-2 L or 20.0 mL
5.00 x 10.3 mol HCI x 1 mol KOH x
mol HCI
0.250 mol KOH
51.
HN0 3(aq) + NaOH(aq)
--+
NaN0laq) + Hp(I)
15.0 g NaOH x I mol NaOH = 0.375 mol NaOH
40.00 g
0.1500 Lx
0.250 mol RN0 3
L
=
0.0375 mol RN0 3
We have added more moles ofNaOH than mol ofRN03 present. Since NaOH and RN0 3 react
in a 1: 1 mol ratio then NaOH is in excess and the solution will be basic. The ions present after
reaction will be the excess OR" ions and the spectator ions, Na+ and N0 3'. The moles of ions
present initially are:
mol NaOH = mol Na+ = mol OH- = 0.375 mol
mol RN03 = mol H+ = mol N03' = 0.0375 mol
CHAPTER 4
74
SOLUTION STOICHIOMETRY
The net ionic reaction occurring is: H+(aq) + OR"(aq) -+ H20(l)
The mol of excess OR" remaining after reaction will be the initial mol ofOH- minus the amount
of OR" neutralized by reaction with H+:
mol excess OR" = 0.375 mol- 0.0375 mol = 0.338 mol OH- excess
The concentration of ions present is:
M
H_
= mol OH - excess = 0.338 mol OH - = 2.25 MOR"
o
volume
-=
M
0.0375 mol N0 3-
N0 3
52.
0.1500L
Ba(OHMaq) + 2 HCI(aq)
75.0
X
0.1500L
-+
= 0.250 MN0 3-; M
= .
0.1500L
Na
= 2.50 MNa+
BaCI2(aq) + 2 HP(I); H+(aq) + OR"(aq)
10-3 L x 0.250 mol HC)
L
225.0 x 10-3 L x
0375 mol
+
= 1.88 x
0.0550 mol Ba(OH)
_=.2
L
10-2 mol HCI
= 1.24
X
= 1.88 x
-+
HP(I)
10-2 mol H+ + 1.88
X
10-2 mol CI­
10-2 mol Ba(OH)2 = 1.24 x 10-2 mol Ba2+
+ 2.48
x
10-2 mol OR"
The net ionic equation requires a 1: 1 mol ratio between OR" and H+. The actual mol OR" to mol
H+ ratio is greater than 1: 1 so OR" is in excess.
Since 1.88 x 10-2 mol OR" will be neutralized by the H+, then we have (2.48 - 1.88)
0.60 x 10-2 mol OR" remaining in excess.
= mol OH - excess
M
total volume
OH -
53.
3
6.0 x 10- mol OH - = 2.0
0.0750 L + 0.2250 L
X
x
10-2 =
10-2 MOR"
HCI(aq) + NaOH(aq) -+ H20(l) + NaCI(aq)
24.16 x 10-3 L NaOH x 0.106 mol NaOH x 1 mol HC) = 2.56 x 10-3 mol HCI
L NaOH
mol NaOH
3
Molarity ofHCI = 2.56 x 10- mol = 0.102 MHCI
25.00 x 10-3 L
35.00
x
10-3 L HN0 3
X
0.0500 mol HN0 3
L HN03
X
1 mol Ca(OH)2
2 mol HN03
X
1 L Ca(OH)2
_
0.0200 mol Ca(OH)2
= 0.0438 L = 43.8 mL Ca(OH)2
CHAPTER 4
55.
Since KHP is a monoprotic acid, the reaction is: NaOH(aq) + KHP(aq)
Mass KHP = 0.02046 L NaOH
56.
75
SOLUTION STOICHIOMETRY
NaOH(aq) + KHP(aq)
0.1082 g KHP
x
---+
0.1000 mol NaOH
L NaOH
x
x
1 mol KHP
mol NaOH
x
---+
H20(l) + NaKP(aq)
204.22 g KHP = 0.4178 g KHP
mol KHP
NaKP(aq) + H20(l)
1 mol KHP
204.22 g KHP
x
1 mol NaOH = 5.298
mol KHP
x
10-4 mol NaOH
There is 5.298 x 10-4 mol of sodium hydroxide in 20.46 mL of solution. Therefore, the
concentration of sodium hydroxide is:
4
5.298 x 10- mol = 2.589 x 10-2 MNaOH
20.46 x 10-3 L
Oxidation-Reduction Reactions
57.
Apply rules in Table 4.2.
a. KMn0 4 is composed ofK+ and Mn04- ions. Assign oxygen a value of -2 which gives
manganese at +7 oxidation state since the sum of oxidation states for all atoms in Mn0 4­
mustequalthe-1 charge on Mn0 4-. K,+l; 0,-2; Mn,+7.
b. Assign
°a -2 oxidation state, which gives nickel a +4 oxidation state. Ni, +4; 0, -2.
c. K4Fe(CN)6 is composed ofK+ cations and Fe(CN)t anions. Fe(CN)64- is composed of iron
and CN- anions. For an overall anion charge of -4, iron must have a +2 oxidation state.
d. (NH4)2HP04 is made ofNH/ cations and HPOt anions. Assign +1 as oxidation state ofH
and -2 as oxidation state ofO. In NH/, x + 4(+1) = +1, x = -3 = oxidation state ofN. In
HPOt, +1 + Y + 4(-2)= -2,y =+5 = oxidation state ofP.
58.
0, -2; Fe, + 8/3
e. 0, -2; P, +3
f.
g. 0, -2; F, -I; Xe, +6
h. F, -I; S, +4
j.
l.
0, -2; C, +2
a.
UO/+: 0, -2; For U, x + 2(-2) = +2, x = +6
Na, +1; 0, -2; C, +3
b. As 20 3 : 0, -2; For As, 2(x) + 3(-2) = 0, x = +3
c. NaBi0 3 : Na, +1; 0, -2; For Bi, +1 + x + 3(-2) = 0, x = +5
e. HAs0 2: assign H = +1 and 0=-2; For As, +1 + x + 2(-2) = 0; x = +3
CHAPTER 4
76
SOLUTION STOICHIOMETRY
Mg, +2; 0, -2; P, +5
g. N~S203: Composed ofNa+ ions and S20t ions.
I.
59.
Na, +1; 0, -2; S, +2
Ca(N03)2: Composed of Ca2+ions and N03- ions.
OCI':
Ca, +2; 0, -2; N, +5
Oxidation state of oxygen is (-2).
-2+x=-1, x=+l;
CI02-: 2(-2) + x = -1, x = +3
The oxidation state of CI in OCI- is + 1.
CI03-: 3(-2) + x = -1, x = +5
b. -3
e. +1
h. +5
CI04-: 4(-2) + x = -1, x = +7
60.
a. -3
d. +2
g. +3
61.
To detennine if the reaction is an oxidation-reduction reaction, assign oxidation numbers. If the
oxidation numbers change for some elements, then the reaction is a redox reaction. If the
oxidation numbers do not change, then the reaction is not a redox reaction. In redox reactions
the species oxidized (called the reducing agent) shows an increase in the oxidation numbers and
the species reduced (called the oxidizing agent) shows a decrease in the oxidation numbers.
Redox?
a.
b.
c.
d.
e.
f.
Yes
Yes
No
Yes
Yes
Yes
c.
f.
i.
2(x)+4(+l)=0, x=-2
+4
0
Oxidizing
Agent
Reducing
Agent
Substance
Oxidized
Substance
Reduced
O2
HCI
CH4
Zn
CH4 (C)
Zn
O 2 (0)
HC1 (H)
03
H20 2
CuCI
NO
H20 2
CuCI
NO(N)
HP2 (0)
CuCI (Cu)
0 3 (0)
H20 2 (0)
CuCI (Cu)
In c, no oxidation numbers change from reactants to products.
62.
Redox?
a.
b.
c.
d.
e.
Yes
No
No
Yes
No
Oxidizing
Agent
Reducing
Agent
Substance
Oxidized
Ag+
Cu
Cu
Ag+
SiCl 4
Mg
Mg
SiCl 4 (Si)
In b, c, and e, no oxidation numbers change.
Substance
Reduced
CHAPTER 4
63.
SOLUTION STOICHIOMETRY
77
Use the method of half-reactions described in Section 4.1 0 of the text to balance these redox
reactions. The first step always is to separate the reaction into the two half-reactions, then
balance each half-reaction separately.
a. Zn -+ Zn 2++ 2 e-
2e- + 2 HCI-+ H2 + 2 CI-
Adding the two balanced half-reactions, Zn(s) + 2 HCI(aq) -+ Hig) + Zn 2+(aq) + 2 CI-(aq)
CIO- -+ CI­
2e- + 2H+ + CIO- -+ CI- + Hp
Adding the two balanced half-reactions so electrons cancel:
N03- -+ NO + 2 Hp
c. As 20 3 -+ H3As04
4 H+ + N03- -+ NO + 2 H20
As 20 3 -+ 2 H 3As04
(3 e- + 4 H+ + N03- -+ NO + 2 HP) x 4
Left 3 - 0; Right 8 Right hand side has 5 extra 0.
Balance the oxygen atoms first using Hp, then balance H using H+, and finally balance
charge using electrons.
°
Common factor is a transfer of 12 eO. Add half-reactions so electrons cancel.
12e-+16H++4N03- -+4NO+8HP
15 Hp + 3 AsP3 -+ 6 H3As04 + 12 H+ + 12 e­
Mn04- -+ Mn 2++ 4 H20
(5 e- + 8 H+ + Mn04 - -+ Mn 2++ 4 H20)
X
2
Common factor is a transfer of 10 e-.
10 Br- -+ 5 Br2 + 10 e­
10 e- + 16 H+ + 2 Mn04- -+ 2 Mn 2++ 8 H20
e.
CH30H -+ CH2 0
(CH30H -+ CHp + 2 H+ + 2 eo)
x
3
Cr20/" -+ Ci3+
14 H+ + Crp/" -+ 2 ci3+ + 7 H20
6 e- + 14 H+ + Cr20 72- -+ 2 Ci3+ + 7 Hp
78
CHAPTER 4
SOLUTION STOICHIOMETRY
Common factor is a transfer of 6 eO.
3 CH30H --+ 3 CH20 + 6 HI- + 6 e­
6 eO + 14 H+ + Cr20/" --+ 2 Cr+ + 7 H 20
64.
a.
(Cu
--+
Cu 2+ + 2 eo)
N0 3­ --+ NO + 2 H 20
(3 e- + 4 H+ + N0 3­ --+ NO + 2 H 20)
3
x
X
2
Adding the two balanced half-reactions so electrons cancel:
3 Cu --+ 3 Cu2+ + 6 e­
6 e- + 8 H+ + 2 N0 3­ --+ 2 NO + 4 H 20
b. (2 Clo
--+
Cl2 + 2 eo)
x
3
Cr20/" --+ 2 Cr+ + 7 H 20
6 e- + 14 H+ + Crp/" --+ 2 Cr+ + 7 Hp
Add the two half-reactions with six electrons transferred:
6 CI- --+ 3 Cl2 + 6 e­
6 eO + 14 H+ + Cr20/" --+ 2 Cr+ + 7 H 20
c.
Pb --+ PbS04
Pb + H2S04 --+ PbS04 + 2 H+
Pb + H 2S04 --+ PbS04 + 2 H+ + 2 e­
Pb0 2 --+ PbS04
Pb02 + H2 S04 --+ PbS0 4 + 2 H 20
2 eO + 2 H+ + Pb02 + H 2 S04 --+ PbS04 + 2 H 20
Add the two half-reactions with two electrons transferred:
2 e- + 2 H+ + Pb0 2 + H 2 S04 --+ PbS04 + 2 H 20
Pb + H 2 S04 --+ PbS04 + 2 H+ + 2 e-
This is the reaction that occurs in an automobile lead storage battery.
d.
Mn 2+ --+ Mn04­
(4 H20 + Mn 2+ --+ Mn04 ­ + 8 H+ + 5 eo)
x
2
CHAPTER 4
79
SOLUTION STOICHIOMETRY
8 Hp + 2 Mn 2+ -+ 2 Mn04- + 16 H+ + 10 e­
10 e- + 30 H+ + 5 NaBi03 -+ 5 Be+ + 5 Na+ + 15 H 20
Simplifying:
e.
(Zn
H 3As0 4 -+ AsH3
H 3As0 4 -+ AsH3 + 4 H 20
8 e- + 8 H+ + H 3As0 4 -+ AsH3 + 4 H 20
-+
Zn 2+ + 2 eo)
x
4
8 e- + 8 H+ + H 3As0 4 -+ AsH3 + 4 H 2 0
4 Zn -+ 4 Zn 2+ + 8 e­
65.
Use the same method as with acidic solutions. After the final balanced equation, then convert H+
to OH" as described in section 4.10 of the text. The extra step involves converting H+ into H 20
by adding equal moles ofOH" to each side of the reaction. This converts the reaction to a basic
solution while keeping it balanced.
a.
Al
-+
AI(OH)4-
4 H 20 + Al -+ AI(OH)4- + 4 H+
4 H 20 + Al -+ AI(OH)4- + 4 H+ + 3 e"
Mn0 4" -+ Mn0 2
3 e- + 4 H+ + Mn0 4" -+ Mn0 2 + 2 H 20
4 H 20 + Al -+ AI(OH)4- + 4 H+ + 3 e"
3 e- + 4 H+ + Mn04" -+ Mn0 2 + 2 H 20
Since H+ doesn't appear in the final balanced reaction, we are done.
b.
Cl 2 -+ CI2 e- + CI 2 -+ 2 CI-
Cl 2 -+ OCI"
2 H 20 + Cl 2 -+ 2 OCl- + 4 H+ + 2 e"
2 e- + Cl2 -+ 2 CI­
2 H 20 + CI 2 -+ 2 OCI- + 4 H+ + 2 e"
Now convert to a basic solution. Add 4 OH- to both sides of the equation. The 4 OR" will
react with the 4 H+ on the product side to give 4 H 20. After this step, cancel identical
species on both sides (2 H 20). Applying these steps gives: 4 OR" + 2 CI 2 -+ 2 CI- + 2 OCl- +
2 H 20, which can be further simplified to:
2 OR(aq) + CI 2(g)
-+
CI-(aq) + OCI-(aq) + Hp(I)
CHAPTER 4
80
c.
SOLUTION STOICHIOMETRY
AI--+ AI0 2"
(2 Hp + AI--+ AI0 2"+ 4 H+ + 3 eo)
N0 2' --+ NH3
6 e' + 7 H+ + N0 2' --+ NH3 + 2 H20
x
2
Common factor is a transfer of 6 eO.
6e" + 7 H+ + N0 2" --+ NH3 + 2 H20
4 H20 + 2 Al --+ 2 AI0 2"+ 8 H+ + 6 e'
66.
a.
Cr --+ Cr(OH)3
crO/" --+ Cr(OH)3
3 H20 + Cr --+ Cr(OH)3 + 3 H+ + 3 e'
3 e' + 5 H+ + CrO/" --+ Cr(OH)3 + H20
3 H20 + Cr --+ Cr(OH)3 + 3 H+ + 3 e"
3 e' + 5 H+ + crO/" --+ Cr(OH)3 + H20
2 OH" + 2 H+ + 2 H20 + Cr + crO/" --+ 2 Cr(OH)3 + 2 OH"
Two OR" were added above to each side to convert to basic solution. The two OH" react with
the 2 H+ on the reactant side to produce 2 H20. The overall balanced equation is:
4 H20(l) + Cr(s) + CrO/"(aq) --+ 2 Cr(OHMs) + 2 OH'(aq)
b.
S2. --+ S
(S2. --+ S + 2 e')
x
5
Mn04" --+ MnS
Mn0 4" + S2" --+ MnS
( 5 e" + 8 H+ + Mn04 ' + S2. --+ MnS + 4 H 20)
X
2
Common factor is a transfer of 10 e'.
5 S2.--+ 5 S+ 10e'
10 e" + 16 H+ + 2 Mn04" + 2 S2" --+ 2 MnS + 8 H 20
Reducing gives: 8 H 20(l) + 7 S2"(aq) + 2 Mn04"(aq) --+ 5 S(s) + 2 MnS(s) + 16 OH"(aq)
c.
CN" --+ CNO'
(H20 + CN" --+ CNO" + 2 H+ + 2 eo)
x
3
Mn04" --+ Mn0 2
(3 e" + 4 H+ + Mn04' --+ Mn0 2 + 2 HP)
x
2
CHAPTER 4
81
SOLUTION STOICHIOMETRY
Common factor is a transfer of 6 electrons.
3 H20 + 3 CN' -. 3 CNO' + 6 H+ + 6 e6 e' + 8 H+ + 2 Mn0 4' - . 2 Mn0 2 + 4 H20
Reducing gives:
H20(l) + 3 CN'(aq) + 2 Mn0 4'(aq) -. 3 CNO'(aq) + 2 MnOb) + 2 OR(aq)
We could balance this reaction by the half-reaction method or by inspection. Let's try
inspection. To balance Cl', we need 4 NaCl:
Balance the Na+ and
sot ions next:
On the left side: 4-H and 10-0; On the right side: 8-0 not counting H20
We need 2 H20 on the right side to balance H and 0:
68.
Au + HN0 3 + HCI -. AuCI4' + NO
Only deal with ions that are reacting (omit H+): Au + N0 3' + CI- -. AuCI 4' + NO
The balanced half-reactions are:
Adding the two balanced half-reactions:
Additional Exercises
69.
4.25 g Cax
Molarity =
1 mol Ca
40.08 g Ca
x
1 mol Ca(OH)2
mol Ca
0.212 mol = 0.942 MOR
225 x 10-3 L
x
2 mol OH mol Ca(OH)2
=0.212molOR
CHAPTER 4
82
70.
mol CaClz present = 0.230 L CaClz x
0.275 mol CaCI
SOLUTION STOICHIOMETRY
2 = 6.33
L CaCI 2
x
10- z mol CaClz
The volume of CaClz solution after evaporation is:
6.33
x
lO- z mol CaClz x
1 L CaCI
2 = 5.75
1.10 mol CaCI 2
X
lO'z L = 57.5 mL CaCl z
Volume HzO evaporated = 230. mL - 57.5 mL = 173 mL HzO evaporated
71.
For the following answers, the balanced molecular equation is first, followed by the complete
ionic equation with the net ionic equation last.
a.
2 AgN03(aq) + BaClz(aq) ---+ 2 AgCI(s) + Ba(N03Maq)
Ag+(aq) + CI'(aq)
---+
AgCI(s)
b. No reaction occurs since all of the possible products (NH4CI and KN0 3) are soluble.
c.
(NH4 )zS(aq) + FeCIz<aq)
---+
FeS(s) + 2 NH4CI(aq)
2 NH/(aq) + SZ-(aq) + Fez+(aq) + 2 CI'(aq)
---+
2 K\aq) + CO/'(aq) + Cuz+(aq) + SOt(aq)
0.0300 L x
0.400 mol Ba(N03)2
L
0.0200 L x
M 2+
Ba
73.
=
excess mol Ba 2+
total volume
---+
1 mol Ba 2+
mol Ba(N03)2
0.200 mol KzC0 3
L
x
FeS(s) + 2 NH/(aq) + 2 CI'(aq)
2-
x
1 mol C03
mol K2C03
x
CuCOb) + 2 K+(aq) + SOt(aq)
= 0.0120 mol Baz+present initially
I mol Ba 2+
= 4.00
x 10'3 mol Baz+reacted
mol CO 23
0.0120 mol - 4.00 x 10-3 mol
0.0200 L + 0.0300 L
8.0 x 10-3 mol
0.0500 L
=
0.16 MBa z+
a.
No; No element shows a change in oxidation state.
b.
1.0 L oxalic acid x 0.14 mol oxalic acid x
1 mol Fe20 3 x 159.70 g Fe20 3
L
6 mol oxalic acid
mol Fe20 3
= 3.7 g Fe 0
Z
3
CHAPTER 4
74.
83
SOLUTION STOICHIOMETRY
M 2 SOiaq) + BaCl 2(aq)
2.33 g BaS04 x
~
BaSOb) + 2 MCI(aq)
I mol BaSO
4
x
I mol M SO
2
233.4 g BaS04
= 9.98
4
mol BaS04
x
10'3 mol M 2S04
From the problem, 1.42 g M 2 S04 was reacted so:
142 amu = 2(atomic mass M) + 32.07 + 4(16.00), atomic mass M = 23 amu
From periodic table, M = Na(sodium).
75.
Use the silver nitrate data to calculate the mol CI' present, then use the formula of douglasite to
convert from CI' to douglasite. The net ionic reaction is: Ag+ + CI' ~ AgCI(s).
0.03720 L x 0.1000 mol Ag + x 1 mol CI - x 1 mol douglasite
L
molAg+
4molClx
311.88 g douglasite =.
0 2900 g doug I't
aSl e
mol
Mass % douglasite = 0.2900 g x 100 = 63.74%
0.4550 g
76.
All the sulfur in BaS04 came from the saccharin. The conversion from BaS04 to saccharin
utilizes the molar masses of each.
0.5032 g BaS04 x
32.07 g S
x 183.19 g saccharin = 0.3949 g saccharin
233.4 g BaS04
32.07 g S
Avg. mass = 0.3949 g = 3.949 x .10-2 g = 39.49 mg
Tablet
10 tablets
tablet
tablet
Avg. mass
77.
0/
1'0
= 0.3949 g saccharin
0.104 g AgCI x
0.5894 g
X
100 =
6700°/
' by mass
. 1'0 saccharm
35.45 gCI- = 2.57 x 10-2 g CI' = Cl' in chlorisondiamine
143.4 gAgCI
Molar mass ofchlorisondiamine = 14(12.01) + 18(1.008) + 6(35.45) + 2(14.01) = 427.00 g/mol
There are 6(35.45) = 212.70 g chlorine for every mole (427.00 g) ofchlorisondiamine.
2.57
X
10'2 g Cl' x
427.00 g drug
212.70gCI-
= 5.16 x
2
10' g drug; % drug =
5 16 x 10-2 g
.
1.28g
x 100 = 4.03%
CHAPTER 4
84
78.
a.
Fe3+(aq) + 3 OH-(aq)
-+
SOLUTION STOICmOMETRY
Fe(OHMs)
Fe(OH)3: 55.85 + 3(16.00) + 3(1.008) = 106.87 glmol
0.107 g Fe(OH)3 x
55.85 g Fe
= 0.0559 g Fe
106.87 g Fe(OH)3
b. Fe(N03)3: 55.85 + 3(14.01) + 9(16.00) = 241.86 glmol
0.0559 g Fe x
c.
79.
241.86 g Fe(NO )
3 3 = 0.242 g Fe(N03)3
55.85 g Fe
Mass % Fe(N03)3 = 0.242 g x 100 = 53.1%
0.456 g
HC 2 H30 2(aq) + NaOH(aq)
a.
-+
H20(l) + NaC 2H30 2(aq)
16.58 x 10-3 L soln x 0.5062 mol NaOH x 1 mol acetic acid = 8.393 x 10-3 mol acetic acid
L soln
mol NaOH
3
. 0 f acetic
. aCl'd = 8.393 x 10- mol = 08393
M
.
ConcentratIon
0.01000 L
b. Ifwe have 1.000 L of solution: total mass = 1000. mL x 1.006 g = 1006 g
mL
Mass of HC 2H30 2 = 0.8393 mol x 60.05 g = 50.40 g
mol
Mass % acetic acid = 50.40 g x 100 = 5.010%
1006g
80.
Mg(s) + 2 HCI(aq)
3.00 g Mg x
81.
--+
MgCI 2(aq) + Hig)
1 mol Mg x 2 mol HCI x 1 L HCI - 0.0494 L = 49.4 mL HCI
24.31 g Mg
mol Mg
5.0 mol HCI
Let HA = unknown acid; HA(aq) + NaOH(aq)
--+
NaA(aq) + H20(l)
mol HA present = 0.0250 Lx 0.500 mol NaOH x 1 mol HA = 0.0125 mol HA
L
I mol NaOH
_x...=g'----HA_ = 2.20 g HA ,x = molar mass ofHA = 176 glmol
mol HA 0.0125 mol HA
Empirical formula weight'" 3(12) + 4(1) + 3(16) = 88 glmol
CHAPTER 4
82.
a.
SOLUTION STOICHIOMETRY
Al(s) + 3 HCI(aq)
~
85
AICI 3(aq) + 3/2 Hig) or 2 AI(s) + 6 HCI(aq)
~
2 AICI 3(aq) + 3 Hig)
Hydrogen is reduced (goes from +1 oxidation state to 0 oxidation state) and aluminum Al is
oxidized (0 ~ +3).
b. Balancing S is most complicated since sulfur is in both products. Balance C and H first then
worry about S.
~
Sulfur is reduced (0
c.
~
-2) and carbon is oxidized (-8/3
~ 2
Silver is reduced (+1
a.
+4).
~
+4).
Although this reaction is mass balanced, it is not charge balanced. We need 2 mol of silver
on each side to balance the charge.
Cu(s) + 2 Ag+(aq)
83.
~
Balance C and H first then balance 0.
Oxygen is reduced (0
d.
-2) and carbon is oxidized (-4
4 NH3(g) + 5 02(g)
-3 +1
~
+2 -2
~
+2).
4 NO(g) + 6 HP(g)
+2 -2
+1 -2 oxidation numbers
--+
2 N0 2(g)
0
+4-2
3 N0 2(g) + H20(l)
+4 -2
0) and copper is oxidized (0
~
0
2 NO(g) + 02(g)
Ag(s) + Cu2+(aq)
~
+1-2
2 HN0 3(aq) + NO(g)
+1 +5-2
+2 -2
All three reactions are oxidation-reduction reactions since there is a change in oxidation
states of some of the elements in each reaction.
b. 4 NH 3 + 5 O 2
~
4 NO + 6 HP; O 2 is the oxidizing agent and NH3 is the reducing agent.
2 NO + O2 ~ 2 N0 2; O 2 is the oxidizing agent and NO is the reducing agent.
3 N0 2 + H20
Mn
~
~ 2
HN03 + NO; N0 2 is both the oxidizing and reducing agent.
Mn 2+ + 2 e"
HN0 3 ~N02
HN0 3 --+ N0 2 + Hp
(e- + H+ + HN0 3 ~ N0 2 + HP) x 2
CHAPTER 4
86
SOLUTION STOICIllOMETRY
Mn ~ Mn 2++ 2 e2 e- + 2 H+ + 2 HN0 3 ---. 2 N02 + 2 H20
8 H20 + 2 Mn 2+ ~ 2 Mn04- + 16 H+ + 10 e10 e- + 10 H+ + 51°4- ---.51°3- + 5 H20
85.
Fe2+will react with Mn0 4- (purple) producing Fe3 +and Mn 2+(almost colorless). There is no
reaction between Mn04- and Fe3+since Fe3+will not oxidize to a higher oxidation state.
Therefore, add a few drops of the potassium permanganate solution. If the purple color persists,
the solution contains iron(III) sulfate. If the color disappears, iron(II) sulfate is present.
Challenge Problems
86.
a.
·
t
5.0 ng Hg
5.0 ppb Hg III wa er =
g soln
5.0 x 10-9 g Hg
mL soln
=: ---~---=
9
5.0 x 10- g Hg x 1 mol Hg x 1000 mL
mL soIn
200.6 g Hg
L
9
b.
c.
1.0 x 10- g CHCI
_ _ _ _ _ _3 x
mL
10.0 ppm As = 10.0 /lg As
g soln
a.
6
x
8.4 x 10-9 M CHCI
3
10.0 x 10- gAs
mL soln
=
1.33 x 10-4 M As
0.10 x 10- g DDT x 1 mol DDT x 1000 mL
mL
354.46 g DDT
L
0.308 g AgCI
=
6
=:
6
87.
2.5 x 10-8 M Hg
1 mol CHCI 3_ x 1000 mL
119.37 g CHCI 3
L
10.0 x 10- g As x 1 mol As x 1000 mL
mL soln
74.92 gAs
L
d.
=
35.45 g CI
143.4 g AgCI
=
= 2.8 x
10-7 M DDT
0.0761 g CI; %Cl = 0.0761 g
0.256 g
100 = 29.7% Cl
x
Cobalt(III) oxide, C020 3 : 2(58.93) + 3(16.00) = 165.86 g/mol
0.145 g C020 3
x
117.86 g Co
165.86 g C020 3
=
0.103 g Co; %Co = 0.103 g
0.416 g
The remainder, 100.0 - (29.7 + 24.8) = 45.5%, is water.
x
100 = 24.8% Co
CHAPTER 4
SOLUTION STOICHIOMETRY
87
Assuming 100.0 g of compound:
45.5 g Hp x
2.016 g H
18.02 g H20
=
5.09 g H; %H =
5.09 g H
x 100 = 5.09% H
100.0 g compound
45.5 g H20
16.00 gO
18.02 g H20
=
40.4 gO; %0 =
40.4 gO
100.0 g compound
x
x
100 = 40.4% 0
The mass percent composition is 24.8% Co, 29.7% Cl, 5.09% H and 40.4% O.
b. Out of 100.0 g of compound, there are:
24.8 g Co
x
5.09 g H x
1 mol
= 0.421 mol Co; 29.7 g Cl
58.93 g Co
1 mol
1.008 gH
=
5.05 mol H; 40.4 gO x
x
1 mol
= 0.838 mol Cl
35.45 g Cl
I mol = 2.53 mol 0
16.00 gO
Dividing all results by 0.421, we get CoCI 2.6Hp.
CoC12·6H20(aq) + 2 NaOH(aq)
~
Co(OHMs) + 2 NaCl(aq) + 6 H20(l)
This is an oxidation-reduction reaction. Thus, we also need to
include an oxidizing agent. The obvious choice is O2 ,
88.
Molar masses: KCl, 39.10 + 35.45
=
AgCI, 107.9 + 35.45
74.55 glmol; KBr, 39.10 + 79.90 = 119.00 glmol
= 143.4 glmol;
AgBr, 107.9 + 79.90 = 187.8 glmol
Let x = number of moles of KCl in mixture and y = number of moles of KBr in mixture.
Since Ag+ + Cl' ~ AgCl and Ag+ + Br' ~ AgBr, then x = moles AgCl and y = moles AgBr.
Setting up two equations:
0.1024 g = 74.55 x + 119.0 y and 0.1889 g = 143.4 x + 187.8 Y
Multiply the first equation by 187.8, and subtract from the second.
119.0
0.1889= 143.4x+187.8y
-0.1616=-117.7x-187.8y
0.0273 = 25.7 x,
x = 1.06 x 10'3 mol KCl
1.06
x
10'3 mol KCl
x
74.55 g KCl = 0.0790 g KCl
molKCl
% KCl = 0.0790 g x 100 = 77.1%, % KBT = 100.0 - 77.1
0.1024 g
= 22.9%
CHAPTER 4
88
2-
89.
0.298 g BaS04 x
96.07 g S04
2-
==
233.4 g BaS04
SOLUTION STOICHIOMETRY
0.123 g
sot;
% sulfate ==
0.123 g S04
0.205 g
==
60.0%
Assume we have 100.0 g ofthe mixture ofNa2 S04 and K2 S04. There is:
60.0 g sot x
I mol == 0.625 mol
96.07 g
sot
There must be 2 x 0.625 == 1.25 mol of + I cations to balance the -2 charge of sot.
Let x == number of moles ofK+ and y == number of moles ofNa+, then x + y == 1.25.
The total mass ofNa+ and K+ must be 40.0 g in the assumed 100.0 g of mixture. Setting up an
equation:
x mol K+ x 39.10 g + y mol Na+ x 22.99 g == 40.0 g
mol
mol
So, we have two equations with two unknowns: x + y == 1.25 and 39.10 x + 22.99 y == 40.0
Since x == 1.25 - y, then 39.10(1.25 - y) + 22.99 y == 40.0
48.9 - 39.10 y + 22.99 y == 40.0, -16.11 Y == -8.9
y
==
0.55 mol Na+ and x == 1.25 - 0.55 == 0.70 mol K+
Therefore:
0.70 mol K+ x I mol ~S04 == 0.35 mol K 2 S04; 0.35 mol K2 S04 x 174.27I g == 61 g K 2SO4
mo
2 molK +
Since we assumed 100.0 g, then the mixture is 61 % K 2S04 and 39% N~S04'
c.
92.
H2 Se(aq) + Ba(OHMaq)
--+
2 H20(l) + BaSe(s)
35.08 mL NaOH x _ _
I_L_ x 2.12 mol NaOH x I mol H 2S04 == 3.72 x 10-2 mol H S0
2
4
1000 mL
L NaOH
2 mol NaOH
2
Molarity == 3.72 x 10- mol x 1000 mol == 3.72 MH S0
4
2
10.00mL
L
CHAPTER 4
93.
a.
89
SOLUTION STOICHIOMETRY
MgO(s) + 2 HCI(aq)
~
MgCI 2(aq) + H20(l)
Mg(OH)ls) + 2 HCI(aq)
~
MgCliaq) + 2 H20(l)
AI(OH)is) + 3 HCI(aq) ~ AICI 3(aq) + 3 H20(l)
b.
Let's calculate the number of moles of HCI neutralized per gram of substance. We can get
these directly from the balanced equations and the molar masses of the substances.
2 mol HCI
I mol MgO = 4.962 x 10-2 mol HCI
40.31 g MgO
g MgO
x
mol MgO
mol
_2_
_HCI
__ x
3.429
== -x
I mol Mg(OH)2
mol Mg(OH)2
58.33 g Mg(OH)2
3_
mol_
HCI
_
_ x
I mol AI(OH)3
g Mg(OH)2
3.846
=-x
78.00 g AI(OH)3
mol AI(OH)3
10-2 mol HCI
_
10-2 mol HCI
_
g AI(OH)3
Therefore, one gram of magnesium oxide would neutralize the most 0.10 MHCI.
94.
We get the empirical formula from the elemental analysis. Out of 100.00 g carminic acid there
are:
53.66 g C
x
I mol C = 4.468 mol C; 4.09 g H x
12.01 g C
42.25 gO
x
1 molO
16.00 g 0
I mol H = 4.06 mol H
1.008 g H
=
2.641 molO
Dividing the moles by the smallest number gives:
4.468
2.641
= 1.692'
4.06
' 2.641
=
1.54
These numbers don't give obvious mol ratios. Let's determine the mol C to mol H ratio.
4.468 = 1.1 0 = l..!4.06
10
So let's try 4.06 = 0.406 as a common factor: 4.468 = 11.0'
10
0.406
4.06
'0.406
=
10.0' 2.641
'0.406
=
6.50
Therefore, C22H20013 is the empirical formula.
We can get molar mass from the titration data.
18.02 x 10-3 L soln x 0.0406 mol NaOH x I mol carminic acid
mol NaOH
L soln
Molar mass =
0.3602 g
7.32 x 10-4 mol
= 492 g
mol
=
7.32 x 10-4 mol carminic acid
CHAPTER 4
90
SOLUTION STOICHIOMETRY
The empirical fonnula mass of CzzHzo013 '" 22(12) + 20(1) + 13(16) = 492 g.
Therefore, the molecular fonnula of canninic acid is also C22H zo 0 13 •
Let H.lA represent citric acid where x is the number of acidic hydrogens. The balanced
neutralization reaction is:
H,A(aq) + x Oa(aq) ---. x Hz0(l) + A'""(aq)
"
molOH reacted
= 0.0372 L x
molOH-
x=-----
mol citric acid
0.1 05 mol OH L
= 3.91
"
"
x 10 3 mol OH
3
3.91 x 10- mol = 3.01
1.30 x 10-3 mol
Therefore, the general acid fonnula for citric acid is H3A meaning that citric acid has three acidic
hydrogens per citric acid molecule (citric acid is a triprotic acid).
96.
a.
HCI(aq) dissociates to H+(aq) + ct'(aq). For simplicity let's use H+ and CI" separately.
H+ ---. Hz
(2 H+ + 2 e" ---. Hz) x 3
Fe ---. HFeCI4
(H+ + 4 CI" + Fe ---. HFeCI 4 + 3 eo) x 2
6 H+ + 6 e' ---. 3 Hz
2 H+ + 8 CI" + 2 Fe ---. 2 HFeCI 4 + 6 e"
8 H+ + 8 ct" + 2 Fe ---. 2 HFeCI 4 + 3 Hz
or
b.
8 HCI(aq) + 2 Fe(s) ---. 2 HFeCliaq) + 3 Hig)
103 ' ---. 13'
3 103 ' ---. 13 '
3 103" ---. 13" + 9 HzO
16 e" + 18 H+ + 3103 ' ---. 13 ' + 9 HzO
16 e" + 18 H+ + 3103" ---. 13" + 9 HzO
24 I" ---. 813" + 16 e'
I" ---. 13'
(3 I" ---. 13" + 2 eO) x 8
CHAPTER 4
91
SOLUTION STOICHIOMETRY
Cr(NCS)/- -. C.-J+ + N03- + CO2 + sot
54 HzO + Cr(NCS)64- -. C.-J+ + 6 N03- + 6 CO2 + 6 SO/- + 108 H+
Charge on left -4. Charge on right = +3 + 6(-1) + 6(-2) + 108(+ 1) = +93. Add 97 e- to the
right, then add the two balanced half-reactions with a common factor of 97 e- transferred.
54 HzO + Cr(NCS)/ -. C.-J+ + 6 NO J- + 6 CO2 + 6 sot + 108 H+ + 97 e97 e- + 97 Ce4+-. 97 Ce3+
+ 6 SOt(aq) + 108 H+(aq)
This is very complicated. A check of the net charge is a good check to see if the equation is
balanced. Left: charge = 97(+4) - 4 = +384. Right: charge = 97(+3) + 3 + 6(-1) + 6(-2) +
108(+1) = +384.
d.
Crl J -. crO/- + 104(16 H20 + Crl 3 -. crO/" + 31°4- + 32 H+ + 27 eo)
x
2
Cl2 - . CI(2 e- + Cl2 - . 2 Cn
x
27
Common factor is a transfer of 54 eO.
54 e- + 27 CI 2 - . 54 CI32 H20 + 2 Crl3 -. 2 crO/" + 6 104- + 64 H+ + 54 e-
Add 64 OR" to both sides and convert 64 H+ into 64 H20.
Reducing gives:
e.
Ce4+-. Ce(OH)J
(e- + 3 H20 + Ce4+-. Ce(OH)3 + 3 H+) x 61
cot + NOJcot + 6 N03'
Fe(CN)64- -. Fe(OH)3 +
Fe(CN)/- -. Fe(OH)J + 6
There are 39 extra ° atoms on right. Add 39 HzO to left, then add 75 H+ to right to balance
H+.
CHAPTER 4
92
39 H20 + Fe(CN)64net charge = -4
-+
SOLUTION STOICHIOMETRY
Fe(OH)3 + 6 cot + 6 N0 3- + 75 H+
net charge = +57
Add 61 e- to the right then add the two balanced half-reactions with a common factor of 61 etransferred.
39 H20 + Fe(CN)t -+ Fe(OH)3 + 6 cot + 6 N03- + 75 H+ + 61 e61 e- + 183 H20 + 61 Ce4 + -+ 61 Ce(OH)3 + 183 H+
Adding 258 OR" to each side then reducing gives:
258 OH'(aq) + Fe(CN)64-(aq) + 61 Ce4+(aq)
f.
Fe(OH)2
(H 20 + Fe(OH)2
-+
-+
-+
61 Ce(OHMs) + Fe(OHMs)
Fe(OH)3
Fe(OH)3 + H+ + eo) x 2
HP2 -+ H20
2 e- + 2 H+ + H20 2 -+ 2 Hp
2 H20 + 2 Fe(OH)2 -+ 2 Fe(OH)3 + 2 H+ + 2 e2 e- + 2 H+ + H20 2 -+ 2 H20
Reducing gives: 2 Fe(OHMs) + H20 2(aq)
97.
-+
2 Fe(OHMs)
First we will calculate the molarity ofNaCI while ignoring the uncertainty.
0.150 g
x
1 mol
58.44g
= 2.57 x
3
10-3 moles; Molarity = 2.57 x 10- mol = 2.57
0.1000L
x
2
10- mol
L
.
ianty
' IS
. =
The maxImum
va Iue fior t h
e mo
0.153g x I mol
2
58.44 g = 2.63-x 10--mol
0.0995 L
L
..
' IS
. =
The mlOlmum
va Iue fior t h
e moianty
2
0.147 g x 1 mol
58.44 g = 2.50
x 10- mol
---0.1005L
L
The range of the NaCI molarity is 0.0250 Mto 0.0263 M or we can express this range as 0.0257
± 0.0007 M