Exercício Resolvido - Quimica Geral (78)

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Chapter 4: Chemical Reactions 
 155 
 
119. (M) Compute the amount of AgNO3 in the solution on hand and the amount of AgNO3 in 
the desired solution. the difference is the amount of AgNO3 that must be added; simply 
convert this amount to a mass. 
 
3
3
3
3
3
33
3
3
3
3
3
3
AgNO g 850.0 
 
AgNO mol1
NO Ag g 9.169
AgNO mmol 1000
AgNO mol 1
AgNO mmol )50.250.7(AgNO mass 
AgNO mmol 50.7
soln mL 1
AgNO mmol 0.0750
mL 100.0 desired AgNOamount 
AgNO mmol 50.2
soln mL 1
AgNO mmol 0.0500
mL 50.00 present AgNOamount 
 
 
120. (E) The balanced equation for the reaction is: S8(s) + 4 Cl2(g) 4 S2Cl2(l) 
 Both “a” and “b” are consistent with the stoichiometry of this equation. Neither bottom row 
box is valid. Box (c) does not account for all the S8, since we started out with 3 molecules, 
but end up with 1 S8 molecule and 4 S2Cl2 molecules. Box (d) shows a yield of 2 S8 
molecules and 8 S2Cl2 molecules so we ended up with more sulfur atoms than we started 
with. This, of course, violates the Law of Conservation of Mass. 
 
121. (D) The pertinent equations are as follows: 
 
3 3 3
C N OH 3 HNCO (g) 
2 2 2 28 HNCO + 6 NO 7 N + 8 CO + 4 H O 
 
The above mole ratios are used to calculate moles of C3N3(OH)3 assuming 1.00 g of NO2. 
3 3 32
3 3 3 2
2 2
3 3 3
3 3 3
3 3 3
1 mol C N (OH)1 mol NO 8 mol HNCO
mass C N (OH) 1.00 g NO 
46.00 g NO 6 mol NO 3 mol HNCO
129.1 g mol C N (OH)
 1.25 g C N (OH)
1 mol C N (OH)
 
 
122. (D) The ammonium dichromate reaction is an example of an internal redox reaction. Both the 
oxidizing agent (Cr2O7
2-
) and the reducing agent (NH4
+
) are found in the compound in the 
correct stoichiometry. The third product is N2(g). 
 
 2 NH4
+ 
 N2 + 8 H
+
 + 6 e
- 
 
8 H
+
 + 6 e
-
 + Cr2O7
2-
 Cr2O3 + 4 H2O 
 
 (NH4)2Cr2O7(s) Cr2O3(s) + 4 H2O(l) + N2(g) 
 
1000 g (NH4)2Cr2O7 4 2 2 7 2 2
4 2 2 7 4 2 2 7 2
1 mol (NH ) Cr O 1 mol N 28.0134 g N
× × ×
252.065 g (NH ) Cr O 1 mol (NH ) Cr O 1 mol N
= 111.1 g N2 
 
Chapter 4: Chemical Reactions 
156 
123. (D) There are many ways one can go about answering this question. We must use all of the 
most concentrated solution and dilute this solution down using the next most concentrated 
solution. Hence, start with 345 mL of 01.29 M then add x mL of the 0.775 M solution. The 
value of x is obtained by solving the following equation. 
 
1.29 M 0.345 L 0.775 M 
1.25 M = 
(0.345 ) L
1.25 M (0.345 ) L = 1.29 M 0.345 L 0.775 M 
043125 + 1.25 0.44505 0.775 Thus, 0.0138 0.475
 0.029 L or 29 mL
x
x
x x
x x x
x
 
 A total of (29 mL + 345 mL) = 374 mL may be prepared this way. 
 
124. (M) 
 balanced equation: FeTiO3 + 2H2SO4 + 4H2O = TiOSO4 + FeSO4·7H2O 
3 3 4 2 4 2
3
3 3 4 2
3
4 2
1 kmol FeTiO 1 kmol FeSO 7 H O 278.018 kg FeSO 7 H O
1.00×10 kg FeTiO × × ×
151.725 kg FeTiO 1 kmol FeTiO 1 kmol FeSO 7 H O 
= 1.83×10 kg FeSO 7 H O 
 
 
125. (M) 
3 2 34 2
2 3 4 2
4 2 4 2
2 3
2 3
2 3
1 kmol Fe O 1 kmol FeSO 7 H O 
mass of Fe O = 1.00×10 kg FeSO 7 H O 
278.018 kg FeSO 7 H O 2 kmol FeSO 7 H O 
159.692 kg Fe O
× = 287 kg kg Fe O 
1 kmol Fe O 
 
 
126. (D) 
(a) 6 CO(NH2)2(l) 6 HNCO(l) + 6 NH3(g) C3N3(NH2)3(l) + 3 CO2(g) 
 
(b)
 
3 3 2 32 2
2 2
2 2 2 2
3 3 2 3
3 3
3 3 2 3
3 3 2
1 kmol C N (NH )1 kmol CO(NH )
mass 100.0 kg CO(NH ) × ×
60.063 kg CO(NH ) 6 kmol CO(NH )
126.121 kg C N (NH )
× 29.4 kg C N (
1 kmol C N (NH )
C N (NH ) = 
84 g actual yield
 
100 g theoretical yield
2 3
NH )
 
127. (M) 
 (a) 2 C3H6(g) + 2 NH3(g) + 3 O2(g) 2 C3H3N(l) + 6 H2O(l) 
 (b) For every kilogram of propylene we get 0.73 kilogram of acrylonitrile; we can also say that 
for every gram of propylene we get 0.73 gram of acrylonitrile. One gram of propylene is 
0.0238 mol of propylene. The corresponding quantity of NH3 is 0.0238 mol or 0.405 g; then 
because NH3 and C3H6 are required in the same molar amount (2:2) for the reaction, 0.405 of a 
kg of NH3 will be required for every 0.73 of a kg of acrylonitrile. To get 1000 kg of 
acrylonitrile we need, by simple proportion, 1000×(0.405)/0.73 = 555 kg NH3.