10.7 g of ammonium chloride was mixed with 6 g of calcium hydroxide and the mixture was heated.

10.7 g of ammonium chloride was mixed with 6 g of calcium hydroxide and the mixture was heated. What gas and in what quantity by its mass and volume will be released?

Let’s implement the solution:
1. Let us write the equation according to the problem statement:
2NH4Cl + Ca (OH) 2 = CaCl2 + 2NH3 + 2H2O – ion exchange, ammonia is released;
2. We make calculations:
M (NH4Cl) = 53.5 g / mol;
MCa (OH) 2 = 74 g / mol;
M (NH3) = 17 g / mol.
3. Calculate the number of moles:
Y (NH4Cl) = m / M = 10.7 / 53.5 = 0.2 mol;
Y Ca (OH) 2 = m / M = 6/74 = 0.08 mol (deficient substance);
Calculations are made for the substance in deficiency.
4. Let’s make the proportion:
0.08 mol of Ca (OH) 2 – X mol (NH3);
-1 mol -2 mol hence, X mol (NH3) = 0.08 * 2/1 = 0.16 mol.
5. Find the mass and volume of ammonia:
m (NH3) = Y * M = 0.16 * 17 = 2.756 g;
V (NH3) = 0.16 * 22.4 = 3.584 l.
Answer: the reaction product ammonia was obtained with a mass of 2.756 g and a volume of 3.584 liters.