Calculate the mass and volume of ammonia that will be released when 20 g of ammonium

Calculate the mass and volume of ammonia that will be released when 20 g of ammonium chloride and 20 g of calcium hydroxide are heated?

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

2NH4Cl + Ca (OH) 2 = 2NH3 + CaCl2 + 2H2O;

Two moles of ammonium react with one mole of calcium hydroxide.

Determine the amount of substance in 20 grams of ammonium chloride and 20 grams of calcium hydroxide.

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

N NH4Cl = 20 / 53.5 = 0.374 mol;

M Ca (OH) 2 = 40 + 16 x 2 + 2 = 74 grams / mol;

N Ca (OH) 2 = 20/74 = 0.27 mol;

0.374 mol of ammonium chloride will react with 0.187 mol of calcium hydroxide to form 0.374 mol of ammonia.

Let’s find its mass and volume:

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

V NH3 = 0.374 x 22.4 = 8.378 liters;

The mass of ammonia will be:

M NH3 = 14 + 3 = 17 grams / mol;

m NH3 = 17 x 0.374 = 6.358 grams;