Chem_171_F10_Q10_Metal_Enth_Key

Chem_171_F10_Q10_Metal_Enth_Key - 1. Chem 171 Quiz 10...

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Chem 171 Quiz 10 Version A (Metal Enthalpy Key) 1. Hydrogen gas produced in the following reaction is collected over water at 296 K; the pressure is 742 mmHg. 2 Al (s) + 6 HCl Æ 2 AlCl 3 (aq) + 3 H 2 (g) What volume of the “wet” gas will be collected in the reaction of 1.50 g Al (s) with excess HCl (aq)? P H2O (296 K) = 21.1 mmHg. Al = 26.98 g/mol. PV = nRT. R =0.0821 L atm / K mol. 760 mmHg = 1 atm. P H2 = 742 mmHg – 21.1 mmHg = 721 mmHg 721 mmHg x 1 atm / 760 mmHg = 0.949 atm Moles of Al = 1.50 x 1 mol Al / 26.98 g = 0.0556 mol Moles of H 2 = 0.0566 mol x 3 mol H 2 / 2 mol Al = 0.0834 mol H 2 V = nRT / P, V = 0.0834 mol x 0.0821 L atm / K mol x 296 K / 0.949 atm = 2.14 L 2. The enthalpy of combustion of C to CO 2 is -393.5 kJ/mol C; the enthalpy of combustion of CO to CO 2 is -283.0 kJ/mol CO: C(s) + O 2 (g) Æ CO 2 (g) (Rx 1) Δ H = - 393.5 kJ CO(g) + ½ O 2 (g) Æ CO 2 (g) (Rx 2) Δ H = - 283.0 kJ Using these data, calculate the enthalpy of combustion of C to CO. C (s) + ½ O 2 (g) Æ CO(g) (Rx 3) Reverse Rx 2 and add it to Rx 1 C(s) + O 2 (g) CO 2 (g) (Rx 1) Δ H = - 393.5 kJ CO 2 (g) CO(g) + ½ O 2 (g) Δ H = 283.0 kJ C (s) + ½ O 2 (g) CO(g) (Rx 3) Δ H = -393.5 kJ + 283.0 kJ = -110.5 kJ Chem 171 Quiz 10 Version B 1. Assume that you are using an open-end manometer filled with mineral oil rather than mercury. What is the gas pressure in the bulb (in mm of mercury) if the level of the
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This note was uploaded on 02/05/2012 for the course 160 171 taught by Professor Staff during the Spring '08 term at Rutgers.

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Chem_171_F10_Q10_Metal_Enth_Key - 1. Chem 171 Quiz 10...

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