Question

What mass of slaked lime would be required to decompose completely 4 grams of ammonium chloride and what would be the mass of each product ?

Answer 1

The reaction for the decomposition of the Slaked lime and Calcium Hydroxide is ⇒

2NH₄Cl + Ca(OH)₂ --------→ CaCl₂ + 2H₂O + 2NH₃↑

Molar mass of the NH₄Cl = 53.5 g/mole.
Molar mass of the Ca(OH)₂ = 74 g/mole.
Molar mass of CaCl₂ = 111
Molar mass of water = 18 g/mole.
Molar mass of Ammonia = 17 g/mole.

From the Reaction,
∵ 2 × 53.5 grams of NH₄Cl requires 74 g of the Ca(OH)₂
∴ 1 g of NH₄Cl requires 74/107 g of the Ca(OH)₂.
∴ 4 g of the NH₄Cl requires 0.692 × 4 g of the Ca(OH)₂.
= 2.768 g of the Ca(OH)₂

Thus, Mass of the Calcium Hydroxide is 2.768 grams.

For the Mass of the CaCl₂.
∵ 2 × 53.5 g of the Ammonium Chloride produces 111 g of Calcium Chloride.
∴ 1 g of the ammonium ch. produces 111/107 g of CaCl₂.
∴ 4 g of NH₄Cl produces 1.04 × 4 g = 4.16 g of CaCl₂

For the mass of water,
107 g of Ammonium chloride produces 2 × 18 g of Water.
∴ 4 g of the Ammonium chloride produces 36/107 × 4 = 1.356 g of water.

For the Mass of Ammonia,
107 g of Ammonium Chloride produces 2 × 17 g of Ammonia.
∴ 1 g of Ammonium Chloride produces 34/107 g of Ammonia.
∴ 4 g of NH₄Cl produces 34/107 × 4 = 1.27 g of NH₃.