Question

Tank of laughing gas comprising 70% mole nitrous oxide (N20) and 30% mole 02.

What is percent by mass of oxygen in a gaseous mixture with molar composition that is 20.00% laughing gas and 80.000% air? Composition of air is 21% mole O2, 79% mole N2, and has average molar mass of 29.0 g/mol

Answer 1

Basis - 100 moles of gaseous mixture

Moles of laughing gas in the mixture = 20% x 100 mol

= 0.20 x 100 = 20 mol

Moles of O2 in laughing gas in the mixture = 30% x 20 mol

= 0.30 x 20 = 6 mol

Mass of O2 = moles x molecular weight

= 6 mol x 32 g/mol

= 192 g

Moles of N2O = 70% x 20 mol

= 0.70 x 20 = 14 mol

Mass of N2O = moles x molecular weight

= 14 mol x 44 g/mol

= 616 g

Mass of laughing gas in mixture = 192 + 616 = 808 g

Moles of air in the mixture = 0.80 x 100 = 80 mol

Mass of air = moles x molecular weight

= 80 mol x 29 g/mol

= 2320 g

Moles of O2 in air in the mixture = 21% x 80

= 0.21 x 80

= 16.8 mol

Total Moles of O2 in the mixture = 6 + 16.8 = 22.8 mol

Mass of O2 in the mixture = moles x molecular weight

= 22.8 mol x 32 g/mol

= 729.6 g

Total mass of mixture = 808 + 2320 = 3128 g

Mass % of O2 in mixture = mass of O2 x 100 / total mass

= 729.6 x 100 / 3128

= 23.32 %