How many grams of ammonium chloride is required to obtain 11.2 liters of ammonia containing 5% impurities

How many grams of ammonium chloride is required to obtain 11.2 liters of ammonia containing 5% impurities, when interacting with calcium hydroxide.

Ca (OH) 2 + 2NH4Cl -> CaCl2 + 2NH3 + 2H2O, find the amount of ammonia = volume / molar. volume = 11.2 / 22.4 = 0.5 mol. from this it follows that the amount of calcium hydroxide is two times less (by coefficient) = 0.25 mol. we find the mass – the number of things * molar = 0.25 * 74 = 18.5 g – this is 90% (since 10% of impurities) we find 100% (we make up the proportion 18.5 / 90 = x / 100 ) it turns out 20.56 g