Question

Aluminum reacts with aqueous sodium hydroxide to produce hydrogen gas according to the following equation: 2Al(s) + 2NaOH(aq) + 6H2O(1)—>2NaAl(OH)4(aq) + 3H2(g) The product gas, H2, is collected over water at a temperature of 20 °C and a pressure of 756 mm Hg. If the wet H2 gas formed occupies a volume of 5.83 L, the number of moles of Al reacted was mol. The vapor pressure of water is 17.5 mm Hg at 20 °C.

Collecting Gas over Water

Answer 1

Write the reacti on as follows 2A1(s) 2NaOH (aq) 2NaAl (OH) (a)+3H2 (g) Given vapor pressure of wateris 17.5 mm Hg and the total pressure is 756 mm Hg total pressure= Partial pressure of oxygen+vapor pressure of water Parti al pressure of H2,A = total pressure - vapor pressure of water 756 mm Hg-17.5 mm Hg = 738.5 mm Hg Therefore, the parti al pressure of H2 is 738.5 mm Hg Volume, V5.83 L Temperature, T 20 °C 273+20 K = 293K gas constant, R = 62.36367 L mmHg/mol K Moles of dry H,,n = RT 738.5 mm Hg x5.83 L 62.36367 L mmHg/mol K x 293K =0.236 mol From the reaction, the stoichiometric relation between Al and H is moles of H moles of Al 2 3 2 (moles of H2) 3 moles of Al = 2 (0.236 mol) = 0.157 mol Therefore, the required solution is 0.157 mol A