Exercício Resolvido - Quimica Geral (70)

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Chapter 4: Chemical Reactions 
 139 
72. (M) The number of grams of TiCl4 that can be made from the reaction mixture is 
determined by finding the limiting reagent, and using the limiting reagent to calculate the 
mass of product that can be formed. The limiting reagent can determined by calculating the 
amount of product formed from each of the reactants. Whichever reactant produces the 
smallest amount of product is the limiting reagent. 
 
 2 4 42 4
2 2 4
1 mol TiO 3 mol TiCl 189.68 g TiCl
35 g TiO 83.1g TiCl
79.88 g TiO 3 mol TiO 1 mol TiCl
 
2 4 4
2 4
2 2 4
1 mol Cl 3 mol TiCl 189.68 g TiCl
45 g Cl 60.2 g TiCl
70.90 g Cl 6 mol Cl 1 mol TiCl
 
2 4 4
4
4
1 mol Cl 3 mol TiCl 189.68 g TiCl
11g C 130 g TiCl
12.01g C 4 mol C 1 mol TiCl
 
 
Cl2 is the limiting reagent. Therefore, 6.0×10
1
 g TiCl4 is expected. 
 
 
Theoretical, Actual, and Percent Yields 
 
73. (M) 
 (a) 44 4
4
1 mol CCl
 277 grams CCl = 1.80 mol CCl
153.81g CCl
 
 Since the stoichiometry indicates that 1 mole CCl F2 2 is produced per mole CCl4 , the 
use of 1.80 mole CCl4 should produce 1.80 mole 2 2CCl F . This is the theoretical yield 
of the reaction. 
 (b) 2 22 2 2 2
2 2
1mol CCl F
187g CCl F = 1.55 mol CCl F
120.91g CCl F
 
The actual yield of the reaction is the amount actually produced, 1.55 mol 2 2CCl F . 
 (c) % % .yield =
1.55mol CCl F obtained
1.80mol CCl F calculated
yield2 2
2 2
100 861% 
 
74. (M) (a) mass C H g C H OH
mol C H OH
g C H OH
mol C H
mol C H OH
6 10 6 11
6 11
6 11
6 10
6 11
= 100.0
1
100.16
1
1
 
6 10
6 10
6 10
82.146g C H
= 82.01g C H = theoretical yield
1mol C H
 
 (b) percent yield =
64.0gC H produced
82.01g C H calculated
yield6 10
6 10
100 78 0%% . 
 (c) mass C H OH g C H produced
g calculated
g produced
mol C H
g C H
6 11 6 10
6 10
6 10
= 100.0
1.000
0.780
1
82.15
 
6 11 6 11
6 11
6 10 6 11
1mol C H OH 100.2g C H OH
= 156g C H OH are needed
1mol C H 1mol C H OH
 
 
 
Chapter 4: Chemical Reactions 
140 
 
75. (D) 
actual yield
% yield = 100%
theoretical yield
 
The actual yield is given in the problem and is equal to 28.2 g. 
In order to determine the theoretical yield, we must find the limiting reagent and do 
stoichiometry. 
 
Conversion pathway approach: 
2 3 3 6 3 6
2 3 3 6
2 3 2 3 3 6
1 mol Al O 2 mol Na AlF 209.94 g Na AlF
7.81g Al O 32.2 g Na AlF
101.96 g Al O 1 mol Al O 1 mol Na AlF
 
3 6 3 6
3 6
3 6
2 mol Na AlF 209.94 g Na AlF0.141 mol
3.50 L 34.5 g Na AlF
1 L 6 mol NaOH 1 mol Na AlF
 
 
Stepwise approach: 
 
 Amount of Na3AlF6 produced from Al2O3 if all Al2O3 reacts 
2 3
2 3 2 3
2 3
3 6
2 3 3 6
2 3
3 6
3 6 3 6
3 6
1 mol Al O
7.81g Al O = 0.0766 mol Al O
101.96 g Al O
2 mol Na AlF
0.0766 mol Al O = 0.153 mol Na AlF
1 mol Al O
209.94 g Na AlF
0.153 mol Na AlF 32.1g Na AlF
1 mol Na AlF
 
Amount of Na3AlF6 produced from NaOH if all NaOH reacts 
3 6
3 6
3 6
3 6 3 6
3 6
0.141 mol NaOH
3.50 L 0.494 mol NaOH
1 L
2 mol Na AlF
0.494 mol NaOH = 0.165 mol Na AlF
6 mol NaOH
209.94 g Na AlF
0.165 mol Na AlF 34.5 g Na AlF
1 mol Na AlF
 
 
Al2O3 is the limiting reagent. 
28.2 g
% yield = 100% 87.6%
32.2 g
 
 
76. (D) The balanced equation is 2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(g) 
 
 
6.63 g
% yield = 100%
theoretical yield
 
3 2 2
3 2
3 3 2
1 mol NH 1 mol N 28.01g N
18.1g NH 14.9 g N
17.03 g NH 2 mol NH 1 mol N