# 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

2NH3    +    H2SO4    →    (NH4)2SO4

34 g                                    132 g

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

∴ 78 g of ammonium sulphate is given by 34\over 132 × 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

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

107 g                                                        34 g

107 g of ammonium chloride gives 34 g of ammonia

∴ 32.6 g of ammonium chloride will give 34\over 107 × 32.6

= 10.36 g of ammonia

Hence, 10.36 g of ammonia gas is obtained.

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