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# First_part - Calculation of Delta Hof for Mg2(aq 1...

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Calculation of Delta H o f for Mg 2+ (aq.) 1. q contents = m contents * C p, contents * Delta T contents q contents = 54.64 g * 3.92 * 13.79 q contents = 2953.66 J q system = - q contents q system = -2953.66 J 2. Mg(s) + 2H + (aq) Mg 2+ (aq) + H 2 (g) Mass of Mg = 0.2051 g Mol of Mg = (0.2051 g)(1 mol / 24.3050 g) = 0.01 mol Mg Mol of H + = (54.64 g)(1 mol / 1.01 g) = 54.10mol H + (0.01 mol Mg)(1 mol Mg 2+ / 1 mol Mg) = 0.01 mol Mg 2+ (54.10mol H + )(1 mol Mg 2+ / 2 mol 2H + ) = 27.05 mol Mg 2+ Since less Mg 2+ is formed with 0.01 mol Mg, Mg is the limiting reactant. Therefore, 0.01 mol Mg 2+ is produced. 3. ΔH 0 rxn = ∑ nΔH 0 f (products) - ∑ mΔH 0 f (reactants) = [ΔH 0 f (H 2 ) + ΔH 0 f (Mg 2+ )] – [ΔH 0 f (Mg) + ΔH 0 f (H + )] = [0 + ΔH 0 f (Mg 2+ )] – [0 + 0] = ΔH 0 f (Mg 2+ ) = Q system / n rxn = -2953.66 / 0.01 mol Mg 2+ = -2953.66/mol

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= -323.692kJ/mol Calculation of Delta H o f for MgO (aq.) 1. q contents = m contents * C p, contents * Delta T contents q contents = 59.88 g *3.92 * 3.97
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## This note was uploaded on 11/11/2010 for the course CHEMISTRY CHEM1800 taught by Professor Krista during the Winter '10 term at UOIT.

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First_part - Calculation of Delta Hof for Mg2(aq 1...

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