Unformatted text preview: ation-reduction reactions? (1) FeSO4 (aq) + KNO3 (aq) + KOH (aq) → K2FeO4 (aq) + KNO2 (aq) + K2SO4 (aq) + H2O (l);
(2) FeSO4 (aq) + KOH (aq) → Fe(OH)2 (s) + K2SO4 (aq);
(3) FeSO4 (aq) + O2 (g) + H2SO4 (aq) → Fe2(SO4)3 + H2O (l);
(4) FeSO4 (aq) + BaCl2 → BaSO4 (s) + FeCl2 (aq);
A) 1 and 2; B) 1 and 3; C) 2 and 3; D) only 1; E) 1, 2 and 3
To answer this question, we need to assign the oxidation numbers to all elements and check if
they are changed. In reaction 1, we observe a change from Fe+2 to Fe+6 in FeO42- anion, as well as
a change from N+5 in NO3- anion to N+3 in NO2- anion. In reactions 2 and 4 there are no changes.
In reaction 3 we have a change from Fe+2 to Fe+3 as well as a change from O20 to O-2 in H2O (l). Chemistry 1024b, Winter 2009 Midterm Exam with Solutions Page 7 out of 9 20. 20.0 ml of aqueous solution of hydrochloric acid was diluted to 100 ml and titrated with a
0.100 M solution of KOH. The equivalence point is reached when 32.0 ml of the latter solution is
added. What is the molarity of the original solution of hydrochloric acid? A) 0.032 M; B) 0.100 M; C) 0.120 M; D) 0.160 M; E) 0.320 M
At the equivalence point, the number of moles nH of H+ (aq) from the acid is exactly the same as
the number of moles nOH of OH- (aq) from the base.
For KOH, nOH = 0.100 mol/litre * 32.0 ml * 0.001 litre/ml = 3.20 * 10-3 mol (3 significant
This means that there were 3.20 * 10-3 mol of HCl (aq) in the original solution. Dilution does not
matter since it improves the titration accuracy but does not change the number of moles of solute.
Hence, the molarity of the original solution was 3.20 * 10-3 mol of solute per 20.0 ml of solution,
or 0.160 M (3 significant figures).
21. Consider the following unbalanced reaction:
Br2 (l) + Cl2 (g) + H2O (l) → HBrO3 (aq) + HCl (aq)
What is the minimum number of moles of water that are required, according to the stoichiometry
of the reaction, to produce 2.0 mol of HCl (aq): A) 0.3; B) 0.6 C) 1.2; D) 2.4; E) 6.0
To answer this question, we need first to balance this oxidation-reduction reaction:
Br20 (l) + 6 H2O (l) – 10 e → 2 Br+5O3-(aq) + 12 H+ (aq);
Cl20 (g) + 2 e → 2 Cl- (aq);
Basically, we have our answer right here, to exchange 10 electrons we need 6 mol...
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