Question

Aluminum reacts with aqueous sodium hydroxide to produce hydrogen gas according to the following equation: 2Al(s) + 2NaOH(aq) + 6H2O(l)2NaAl(OH)4(aq) + 3H2(g) The product gas, H2, is collected over water at a temperature of 20 °C and a pressure of 750 mm Hg. If the wet H2 gas formed occupies a volume of 9.03 L, the number of grams of H2 formed is g. The vapor pressure of water is 17.5 mm Hg at 20 °C.

Answer 1

The partial pressure of H2 will be the total pressure minus the vapor pressure of water: 750 - 17.5 = 732.5 mm Hg

Use the ideal gas law to calculate the moles of H2

P = 732.5/760 = 0.963 atm
T = 20+273 = 293 K
V = 9.03 L
R = 0.0821 L atm/mol K

PV = n RT
(0.963)x(9.03) = n (0.0821)x(293)
n = 0.361 moles H2

moles = mass/molecular weight

mass of H₂ = molecular weight x moles = 2.01 x 0.361

= 0.722 g

The no. of grams of H₂ formed is 0.722 g.

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