What is the volume of nitrogen formed during the combustion of ethylamine weighing 10 g?

Find the amount of ethylamine substance (C2H5NH2).

n = m: M.

M (C2H5NH2) = 45 g / mol.

n = 10 g: 45 g / mol = 0.22 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

4C2H5NH2 + 15O2 = 8СО2 ↑ + 6H2O + 2N2 ↑.

4 mol of ethylamine accounts for 2 mol of nitrogen. The substances are in quantitative ratios 4: 2 = 2: 1. The amount of nitrogen will be 2 times less than the amount of ethylamine.

n (N2) = ½ n (C2H5NH2) = 0.22: 2 = 0.11 mol.

Let’s find the volume of nitrogen.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 22.4 L / mol × 0.11 mol = 2.46 L.

Answer: V = 2.46 liters.