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prelim1_answers - CHEMISTRY 2090 PRELIM I ANSWERS October 7...

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CHEMISTRY 2090 PRELIM I ANSWERS October 7, 2008 1
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1. (a) (8 points) White phosphorus, P 4 , will spontaneously burst into flames if stored in air. It can be neutralized using nitric acid, HNO 3 ; balance the following equation for the reaction in acidic aqueous solution. P 4 (s) + NO 3 (aq) −→ H 2 PO 4 (aq) + NO(g) Divide this one into half-reactions, one for P, the other focusing on N, as follows: P 4 ( s ) + 16H 2 O −→ 4H 2 PO 4 ( aq ) + 24H + ( aq ) + 20e and 4H + ( aq ) + NO 3 ( aq ) + 3e −→ NO ( g ) + 2H 2 O We need to conserve charge, that means we require that the numbers of electrons taken from P equal the number added to the N atoms. So we need to multiply the top reaction 3-fold, and the bottom reaction 20-fold, and add; the result is 3P 4 ( s ) + 48H 2 O + 80H + ( aq ) + 20NO 3 ( aq ) + 60e −→ 12H 2 PO 4 ( aq ) + 72H + ( aq ) + 60e + 20NO ( g ) + 40H 2 O or, eliminating the common reactants and products, 3P 4 ( s ) + 8H 2 O + 8H + ( aq ) + 20NO 3 ( aq ) −→ 12H 2 PO 4 ( aq ) + 20NO ( g ) (b) (8 points) A solution of permanganate (MnO 4 ) is standardized by titration with oxalic acid (H 2 C 2 O 4 ). It required 29.0 m of the permanganate solution to react completely with 0.106 g of the oxalic acid. The unbalanced equation for the reaction is MnO 4 (aq) + H 2 C 2 O 4 (aq) −→ Mn 2+ (aq) + CO 2 (g) What is the molarity of the permanganate solution? We start by balancing the reaction equation. I’ll assume in acid though it really won’t matter. For the manganese half-reaction MnO 4 ( aq ) + 8H + ( aq ) + 5e −→ Mn 2 + ( aq ) + 4H 2 O and for the oxalate half-reaction H 2 C 2 O 4 ( aq ) −→ 2H + ( aq ) + 2CO 2 ( g ) + 2e To balance the pair of couples so that there will be no electrons created or destroyed 2MnO 4 ( aq ) + 6H + ( aq ) + 5H 2 C 2 O 4 ( aq ) −→ 2Mn 2 + ( aq ) + 10CO 2 + 8H 2 O so that 2 MnO 4 ions react with 5 H 2 C 2 O 4 molecules. Now we can find the moles of permanganate required; 0 . 106 g H 2 C 2 O 4 × 1 mole H 2 C 2 O 4 90 . 03 g H 2 C 2 O 4 × 2 mole MnO 4 5 moles H 2 C 2 O 4 = 4 . 709 × 10 4 moles MnO 4 The molarity of the permanganate solution is therefore 4 . 709 × 10 4 moles MnO 4 0 . 0290 solution = 0 . 0162M 2
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2. (a) (9 points) Faraday told us that the mass-to-charge ratio of metals deposited by elec- trolysis is proportional to the current passed through the solution. In an electrolysis experiment a current I 0 is passed through a solution of AgNO 3 for 10.0 minutes, and releases 0.432 g of Ag metal. The same current is passed through a second solution for the same time period. Our second solution contained the plumbite ion, PbO y x , where x the number of oxygen atoms in the plumbite ion–and y –the net charge on the plumbite ion–are to be determined. The relevant (unbalanced) reaction at the electrode is PbO y x (aq) + H 2 O + e −→ Pb(s) + OH (aq) and is found to deposit 0.414 g of metallic Pb, while releasing 8 . 00 × 10 3 moles of OH into solution.
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