The interaction of hydrogen and nitrogen yielded 3.4 g of ammonia, which is 25% of the theoretically possible yield

The interaction of hydrogen and nitrogen yielded 3.4 g of ammonia, which is 25% of the theoretically possible yield. Determine the volume of air containing the volume of nitrogen required for the reaction.

N2 + 3H2 = 2NH3

Let’s find the amount of ammonia substance:

n (NH3) = m (NH3) / M (NH3) = 3.4 / 17 = 0.2 mol;

The theoretically possible amount of ammonia will be:

ntheor (NH3) = n (NH3) / η (NH3) = 0.2 / 0.25 = 0.8 mol;

According to the stoichiometry of the reaction:

n (N2) = 0.5ntheor (NH3) = 0.5 * 0.8 = 0.4 mol;

Let us find the volumes of nitrogen and air (at standard conditions), taking into account that the volume fraction (x) of nitrogen in air is 0.78:

V (N2) = n (N2) * VM = 0.4 * 22.4 = 8.96 L;

V (air) = V (N2) / x (N2) = 8.96 / 0.78 = 11.49 liters.

Answer: V (air) = 11.49 liters.