Find the mass and volume a) 0.5 mol of nitrogen b) 2 mol of carbon dioxide c) 0.3 mol of oxygen

Data: νN2 (amount of nitrogen substance) = 0.5 mol; νСО2 (amount of things carbon dioxide) = 2 mol; νO2 (amount of oxygen) = 0.3 mol.

Const: MN2 – molar mass of nitrogen (MN2 ≈ 28 g / mol); MСО2 – molar mass of carbon dioxide (MСО2 ≈ 44 g / mol); MО2 – molar mass of oxygen (MО2 ≈ 32 g / mol); Vm – molar volume (assumed n.v. and Vm = 22.4 l / mol).

1) The mass of the taken gases:

– nitrogen: mN2 = MN2 * νN2 = 28 * 0.5 = 14 g;

– carbon dioxide: mСО2 = MСО2 * νСО2 = 44 * 2 = 88 g;

– oxygen: mО2 = MО2 * νО2 = 32 * 0.3 = 9.6 g.

2) Volumes of gases taken:

– nitrogen: VN2 = Vm * νN2 = 22.4 * 0.5 = 11.2 l;

– carbon dioxide: VСО2 = Vm * νСО2 = 22.4 * 2 = 44.8 l;

– oxygen: VO2 = Vm * νO2 = 22.4 * 0.3 = 6.72 g.

Answer: The mass of nitrogen is 14 g, the volume is 11.2 liters; mass of carbon dioxide 88 g, volume 44.8 liters; oxygen mass 9.6 g, volume 6.72 g.