What is the volume of ammonia formed when interacting with 53.5 g of ammonium chloride with sodium hydroxide

The reaction of ammonium chloride with sodium hydroxide is described by the following chemical reaction equation:

NH4Cl + NaOH = NH3 + NaCl + H2O;

One mole of ammonium chloride reacts with one mole of sodium hydroxide. This produces one mole of ammonia.

Determine the amount of substance in 53.5 grams of ammonium chloride:

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

N NH4Cl = 53.5 / 53.5 = 1 mol;

Consequently, during the reaction, 1 mol of ammonia was formed.

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

N NH3 = 22.4 liters;