The hydrogen gas formed in a chemical reaction is collected over water at 30.0 °C at a total pressure of 732 mmHg. What is the partial pressure of the hydrogen gas collected in this way? If the total volume of gas collected is 722 mL, what mass of hydrogen gas is collected?

Explanation: i) P2 = 732 mmHg; T1 = 273 K; T2 = (273 + 30) K = 300 K;

∴ P1/T1 = P2/T2

P1/273 = 732/300

P1 = 732 X 273/300 = 666.12 mmHg

∴ Partial pressure = (732 - 666.12) = 65.88 mmHg

ii) Pressure (at atmosphere) = 732/760 = 0.96 atm; R = 0.0821 atm/mole. K; T = 300 K; V = 722/1000 = 0.722 L

∴ Number of moles, n = PV/RT (from nRT = PV)

n = 0.96 X 0.722 / (0.0821 X 300) = 0.6954/24.63 = 0.028 mol

∴ mass of hydrogen gas = Molar mass X n (where molar mass of hydrogen is 2g) = 0.028 X 2 = 0.056g