Exercício Resolvido - Quimica Geral (93)

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Chapter 5: Introduction to Reactions in Aqueous Solutions 
 185 
 
2 3 2 3 2 2 3 2 2 2Na CO s + 2 HC H O aq 2 NaC H O aq + H O(l) + CO g 
 
acetic acid required for solid: 
 
2 3 2 3 2
2 3 2 3 2
2 3 2 3
1000 mmol Na CO 2 mmol HC H O
= 7.55 g Na CO = 143 mmol HC H O
106.0 g Na CO 1 mmol Na CO
 
 
Clearly there is not enough acetic acid present to react with all of the sodium carbonate. The 
resulting solution will not be acidic. In fact, the solution will contain only a trace amount of 
acetic acid (HC2H3O2). 
 
59. (M) 2 3 2 2 3 2base
2 3 2
4.0 g HC H O 1 mol HC H O1.01 g vinegar
V = 5.00 mL vinegar
1 mL 100.0 g vinegar 60.0 g HC H O
 
 
1
1
1
0.1000
1000
1
= 34
2 3 2
 mol NaOH
 mol HC H O
 L base
 mol NaOH
 mL
 L
 mL base 
 
60. (M) The titration reaction is . 2 4 2 4 22 NaOH aq + H SO aq Na SO aq + 2 H O(l) 
It is most convenient to consider molarity as millimoles per milliliter when solving this problem. 
 
H SO
49.74 mL base
0.935mmol NaOH
1mL base
1mmol H SO
2 mmol NaOH
 mL battery acid
 M
2 4
2 4 =
5.00
= 4.65 H2SO4 
Thus, the battery acid is not sufficiently concentrated. 
 
61. (E) Answer is (d): 120 % of necessary titrant added in titration of NH3 
 
 
5 NH3
+
5 HCl
+
1 HCl
required for
equivalence
point
20 % excess
5 NH4
+ + 6 Cl- + H3O
+
(depicted in question's drawing ) 
 
 
62. (E) 
 (a) H2O(l) + K
+(aq) + Cl-(aq) 
 
 (b) CH3COOH(aq) + CH3COO
-(aq) + H2O(l) + Na
+(aq) 
 
Chapter 5: Introduction to Reactions in Aqueous Solutions 
186 
Stoichiometry of Oxidation–Reduction Reactions 
 
63. (M)
 
Conversion pathway approach: 
2 3 4 4
2 3
2 3 2 3 4
4
1mol As O 4 mol MnO 1mol KMnO
0.1078 g As O
197.84 g As O 5 mol As O 1mol MnO
[ MnO ]=
1L
22.15 mL
1000 mL
4
= 0.01968 M KMnO
 
Stepwise approach: 
4
-4 -44
4
4
-42 3
2 3 2 3
2 3
-4 -44
2 3 4
2 3
4
[ KMnO ]=
1mol KMnO
10 mol MnO 10 
1mol MnO
mol KMnO
L solution
1mol As O
0.1078 g As O = 5.449 10 mol As O
197.84g As O
4 mol MnO
5.449 10 mol As O = 4.359 10 mol MnO
5mol As O
4.359 = 4.359
-4
-24 4
4
4mol KMnO
1L
22.15 mL 0.02215 L solution
1000 mL
mol KMnO 4.359 10 mol KMnO
[ KMnO ]= = = 1.968 10 M
L solution 0.02215 L solution
 
 
64. (E) The balanced equation for the titration is: 
2 + 2 2+
3 4 4 25 SO aq + 2 MnO aq + 6 H aq 5 SO aq + 2 Mn aq + 3 H O l 
2
34 4
2 24 4
3 32
3
5 mmolSO0.02237 mmol KMnO 1mmol MnO
31.46 mL
1mL soln 1 mmol KMnO 2 mmol MnO
SO = = 0.07038 M SO
25.00 mL SO soln
 
65. (M) First, we will determine the mass of Fe, then the percentage of iron in the ore. 
2 2+
2 7
2 2+
2 7
0.05051 mol Cr O1 L 6 mol Fe 55.85 g Fe
mass Fe = 28.72 mL
1000 mL 1 L soln 1 mol Fe1 mol Cr O
 
mass Fe = 0.4861 g Fe % .Fe =
0.4861g Fe
0.9132 g ore
Fe100% 53 23%