Exercício Resolvido - Quimica Geral (68)

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Chapter 4: Chemical Reactions 
 135 
57. (M) The volume of solution determines the amount of product. 
 
3 2 4 2 4
2 4
3 2 4
2 4 2 4
0.186mol AgNO 1mol Ag CrO 331.73g Ag CrO1L
mass Ag CrO = 415mL
1000mL 1L soln 2mol AgNO 1mol Ag CrO
mass Ag CrO = 12.8 g Ag CrO
 
58. (M) 
4
4 4
KMnO 4
4
2 mol KMnO 1 L KMnO1 mol KI
V 9.13 g KI = 0.138 L KMnO
166.0023 g KI 10 mol KI 0.0797 mol KMnO
 
 
59. (M) 
1L 0.175mol NaOH 2mol Na 22.99g Na
mass Na =155mL soln
1000mL 1L soln 2mol NaOH 1mol Na
 = 0.624g Na 
 
60. (M) We determine the amount of HCl present initially, and the amount desired. 
 amount HCl present mL
mmol HCl
mL soln
mmol HCl= 250.0
1.023
1
= 255.8 
 amount HCl desired mL
mmol HCl
mL soln
mmol HCl= 250.0
1.000
1
= 250.0 
 
1mmol Mg 24.3mg Mg
mass Mg = 255.8 250.0 mmol HCl = 70. mg Mg
2mmol HCl 1mmol Mg
 
61. (M) The mass of oxalic acid enables us to determine the amount of NaOH in the solution. 
 2 2 4 2 2 4
2 2 4 2 2 4
0.3126g H C O 1000mL 1mol H C O 2mol NaOH
NaOH = = 0.2649M
26.21mL soln 1L soln 90.04g H C O 1mol H C O
 
 
62. (D) The total amount of HCl present is the amount that reacted with the CaCO3 plus the 
amount that reacted with the Ba(OH)2 (aq). 
 
 3
3
3
3 3
moles HClfrom 1mol CaCO 2mol HCl 1000mmol
= 0.1000g CaCO
CaCO reaction 100.09g CaCO 1mol CaCO 1mol
= 1.998mmol HCl
 
 
2
2 2
moles HCl from 0.01185mmol Ba OH 2 mmol HCl
= 43.82 mL
Ba OH reaction 1mL soln 1mmol Ba OH
= 1.039 mmol HCl
 
 The HCl molarity is this total mmol of HCl divided by the total volume of 25.00 mL. 
 [HCl] 
1.998 +1.039 mmol HCl
= = 0.1215M
25.00mL
 
 
 
 
 
Chapter 4: Chemical Reactions 
136 
 
Determining the Limiting Reactant 
 
63. (E) The limiting reactant is NH3 . For every mole of NH3(g) that reacts, a mole of NO(g) 
forms. Since 3.00 moles of NH3(g) reacts, 3.00 moles of NO(g) forms (1:1 mole ratio). 
 
64. (E) The reaction of interest is: CaH2(s) + 2 H2O(l) Ca(OH)2(s) + 2 H2(g) 
 The limiting reactant is H2O(l). The mole ratio between water and hydrogen gas is 1:1. 
 Hence, if 1.54 moles of H2O(l) reacts, 1.54 moles of H2(g) forms (1:1 mole ratio). 
 
65. (M) First we must determine the number of moles of NO produced by each reactant. The 
one producing the smaller amount of NO is the limiting reactant. 
2mol NO
mol NO = 0.696mol Cu = 0.464mol NO
3mol Cu
 
 
Conversion pathway approach: 
3
3
3
6.0 mol HNO1L 2 mol NO
mol NO = 136 mL HNO aq × × × = 0.204 mol NO
1000 mL 1L 8 mol HNO
 
 Stepwise approach: 
 
 
3 3
3
3
3
3
1L
136 mL HNO aq × 0.136 L HNO
1000 mL
6.0 mol HNO
0.136 L × 0.816 mol HNO
1L
2 mol NO
0.816 mol HNO × = 0.204 mol NO
8 mol HNO
 
 
Since HNO3(aq) is the limiting reactant, it will be completely consumed, leaving some Cu 
unreacted. 
 
 
66. (M) First determine the mass of H2 produced from each of the reactants. The smaller mass 
is that produced by the limiting reactant, which is the mass that should be produced. 
 2 2
2 2
2
(Al)
3mol H 2.016g H1mol Al
mass H = 1.84g Al = 0.206g H
26.98g Al 2mol Al 1mol H
 
 2 2
2 2
2
(HCl)
3mol H 2.016g H2.95mol HCl1L
mass H = 75.0mL = 0.223g H
1000mL 1L 6mol HCl 1mol H
 
 Thus, 0.206 g H2 should be produced.