What volume of ammonia is formed by the interaction of Hydrogen with 12.6 g of nitrogen containing 5% impurities?

Let’s find the mass of pure nitrogen without impurities.

100% – 5% = 95%.

12.6g – 100%,

x – 95%,

x = (12.6 g × 95%): 100% = 11.97 g.

Let’s find the amount of nitrogen substance.

n = m: M.

M (N2) = 14 × 2 = 28 g / mol.

n = 11.97 g: 28 g / mol = 0.43 mol.

Let us compose the reaction equation, find the quantitative relations of substances.

3H2 + N2 = 2NH3.

For 1 mole of nitrogen, there are 2 moles of ammonia. Substances are in quantitative ratios 3: 2 = 1.5: 1. The amount of ammonia is 1.5 times less than the amount of hydrogen.

n (NH3) = 0.43: 1.5 = 0.27 mol.

Let’s find the volume of ammonia.

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 × 0.27 mol = 6.048 L.

Answer: V = 6.048 l.