Calculate the weight of ammonia gas.

(a) Required for reacting with sulphuric acid to give 78 g. of fertilizer ammonium sulphate.

(b) Obtained when 32.6 g. of ammonium chloride reacts with calcium hydroxide during the laboratory preparation of ammonia.

[2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃]

[N = 14, H = 1, O = 16, S = 32, Cl = 35.5].

(a) Molecular Weight of Ammonia (NH₃) = 14 + 3(1) = 17 g

Molecular Weight of Ammonium Sulphate ((NH₄)₂SO₄) = 2[14 + 4(1)] + 32 + 4(16) = 132 g

$\underset{2(17) = 34 \text{ g}}{2\text{NH}$3} + \text{H}2\text{SO}4 \longrightarrow \underset{132 \text{ g}}{(\text{NH}4)2\text{SO}4}

132 g of ammonium sulphate is given by 34 g of ammonia

∴ 78 g of ammonium sulphate is given by $\dfrac{34}{132}$ x 78 = 20.09 g of ammonia

Hence, 20.09 g of ammonia gas is required.

(b) Molecular Weight of Ammonium Chloride (NH₄Cl) = 14 + 4(1) + 35.5 = 53.5 g

$\underset{2(53.5) = 107 \text{ g}}{2\text{NH}$4\text{Cl}} + \text{Ca}(\text{OH})2 \longrightarrow \text{CaCl}2 + 2\text{H}2\text{O} + \underset{2(17) = 34 \text{ g}}{2\text{NH}_3}

107 g of ammonium chloride gives 34 g of ammonia

∴ 32.6 g of ammonium chloride will give $\dfrac{34}{107}$ x 32.6 = 10.36 g of ammonia

Hence, 10.36 g of ammonia gas is obtained.