Question

Electrical energy can be used to separate water into O2(g) and H2(g). In one demonstration of this reaction, 41.00 mL of H2 are collected over water at 25°C. Atmospheric pressure is 755.0 mmHg. How many grams of H2 are collected?

Answer 1

Volume of H2 collected V = 41 mL x 1L/1000 mL

= 0.041 L

Temperature T = 25 + 273 = 298 K

Atmospheric pressure = 755.0 mmHg

Vapor pressure of water at 25 C = 23.8 mmHg

Pressure of dry H2

P = Atmospheric pressure - Vapor pressure

= 755 - 23.8

= 731.2 mmHg x 1atm/760mmHg

= 0.9621 atm

From the ideal gas equation

Moles of H2 gas collected

n = PV/RT

= (0.9621 atm x 0.041 L) / (0.0821 L-atm/mol-K x 298K)

= 0.0016123 mol

Mass of H2 gas collected = moles x molecular weight

= 0.0016123 mol x 2g/mol

= 0.003225 g