Exercício Resolvido - Quimica Geral (101)

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Chapter 5: Introduction to Reactions in Aqueous Solutions 
 201 
2 2 4 2 2 2 4 2
2 2 4 2 2 4 2
1 mol H C O 2H O 126.07 g H C O 2H O
1 mol H C O 1 mol H C O 2H O
2 2 4 2= 0.9474 g H C O 2H O 
The mass of H C O H O2 2 4 22 that reacted with MnO2 2 2 4 2= 1.651 g 0.9474 g = 0.704 g H C O 2H O 
 
2 2 4 2 2
2 2 2 4 2
2 2 4 2 2 2 4 2
2
1 mol H C O 1 mol MnO 86.9 g MnO
mass MnO = 0.704 g H C O 2H O
126.07 g H C O 2H O 1 mol H C O 1 mol MnO
= 0.485 g MnO
% MnO
0.485g MnO
0.533gsample
100% 91.0% MnO2
2
2 
 
96. (D) Reactions: 2 NH3 + H2SO4 (NH4)2SO4 and H2SO4 + 2NaOH 2 H2O + Na2SO4 
 
-
2 4
-
-
2 4OH
- 2 4
H SO 2 4-
2 4OH
n (used to find moles H SO in excess) 0.03224 L 0.4498 M = 0.01450 mol OH
1 mol H SO
n ( in excess) = 0.01450 mol OH 0.00725108 mol H SO
2 mol OH
n (used to find moles H SO in separate 
2 4
-
- 2 4
H SO 2 4-
unreacted sample) 0.02224 L 0.4498 M 
 = 0.0100035 mol OH
1 mol H SO
n (initial) = 0.0100035 mol OH 0.005002 mol H SO
2 mol OH
NOTE: this was in a 25.00 mL sample: we need to scale up to 5
2 4
2 4 2 4 2 4
3
H SO 2 4 2 4
H SO H SO H SO
2 4 2 4 2 4
NH
0.00 mL.
Hence, n (initial) = 2 0.005002 mol H SO = 0.0100035 mol H SO 
n (reacted) = n (initial) n (excess) 
= 0.0100035 mol H SO 0.00725108 mol H SO = 0.00275 mol H SO
n 0.
3
3
2 4 3
2 4
NH 3
3
2 mol NH
00275 mol H SO 0.005505 mol NH
1 mol H SO
1 mol N 14.0067 g N
mass 0.005505 mol NH 0.0771 g N in sample
1 mol NH 1 mol N
100 g protein
mass protein in sample = 0.0771 g N in sample
16 
= 0.482 g protein in sample
g N
0.482 g protein in sample
percent protein in sample = 100% = 38.6 % protein
1.250 g sample
 
 
97. (D) 
 The molecular formula for CH3CH2OH is C2H6O and for CH3COOH is C2H4O2. 
The first step is to balance the oxidation–reduction reaction. 
Oxidation: [C2H6O + H2O C2H4O2 + 4 H
+ + 4 e ] × 3 
Reduction: [Cr2O7
2 + 14 H+ + 6e 2 Cr3+ + 7 H2O] × 2 
Overall: 3 C2H6O + 2 Cr2O7
2 + 16 H+ 3 C2H4O2 + 4 Cr
3+ + 11 H2O 
Chapter 5: Introduction to Reactions in Aqueous Solutions 
202 
Before the breath test: 
-42 2 70.75 mg K Cr O 1 g 1 mol 1000 mL = 8.498 10 M
3 mL 1000 mg 294.19 g 1 L
 
 = 8×10-4 M (to 1 sig fig) 
For the breath sample: 
BrAC = 2 6
0.05 g C H O 1 mL blood
100 mL blood 2100 mL breath
 = 
7
2 62.38×10 g C H O
mL breath
 
mass C2H6O = 
7
2 62.38×10 g C H O
mL breath
 × 500. mL breath = 1.19 × 10 4 g C2H6O 
 
Calculate the amount of K2Cr2O7 that reacts: 
–
–
2
4 2 6 2 7 2 2 7
2 6 2
2 6 2 6 2 7
6
2 2 7
1 mol C H O 2 mol Cr O 1 mol K Cr O
1.19 10 g C H O 
46.068 g C H O 3 mol C H O 1 mol Cr O
= 1.72 10 mol K Cr O
 
 
# mol K2Cr2O7 remaining = moles K2Cr2O7 before – moles K2Cr2O7 that reacts 
moles K2Cr2O7 before = 
-6
2 2 7
1 g 1 mol
0.75 mg K Cr O = 2.5 10 mol
1000 mg 294.19 g
 
# mol K2Cr2O7 remaining = 2.5 × 10
6 mol 1.72 × 10 6 mol = 0.78 × 10 6 mol 
 
concentration of K2Cr2O7 after the 
 breath test = 0.78 × 10 6 mol/0.003 L = 2.6 × 10 4 mol/L = 3 × 10 4 mol/L (to 1 sig fig) 
 
98. (D) 
 (a) Step 1: Assign oxidation states to each element in the reaction and identify the species 
being oxidized and reduced. 
The oxidation state of Cr is +6 in Cr2O7
2 and +3 in Cr3+. The oxidation state of Cl is 
1 in Cl and 0 in Cl2. Each Cr gains three electrons and each Cl loses one electron. 
Step 2: Write separate, unbalanced equations for the oxidation and reduction half-reactions. 
 Oxidation: Cl Cl2 
 Reduction: 22 7Cr O Cr
3+