In the interaction of 2.24 liters of ammonia with hydrochloric acid, 5 g of ammonium chloride was obtained.

In the interaction of 2.24 liters of ammonia with hydrochloric acid, 5 g of ammonium chloride was obtained. Calculate the mass fraction of the yield of ammonium chloride from the theoretically possible.

The reaction of interaction of ammonia with hydrochloric acid is described by the following chemical reaction equation:

NH3 + HCl = NH4Cl;

In this reaction, one mole of ammonium chloride is formed from one mole of ammonia and hydrochloric acid.

Determine the amount of substance in 5 grams of ammonium chloride.

M NH4Cl = 14 + 4 + 35.5 = 53.5 grams / mol;

N NH4Cl = 5 / 53.5 = 0.093 mol;

The theoretically possible amount of salt is equal to the amount of ammonia substance

Under normal conditions, one mole of ideal gas takes up a volume of 22.4 liters.

The required amount of ammonia will be:

N NH3 = 2.24 x 22.4 = 0.1 mol;

The product yield will be equal to the ratio of the practical and theoretical amount of the substance and will be:

K = 0.093 / 0.1 = 0.93 = 93%;