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The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by ${NiCl}_{2} . 6 H_{2} O$ to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952 g of ${NiCl}_{2} . 6 H_{2} O$ are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is (Atomic weights in g mol-1: H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59)

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answer is 2992.

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Detailed Solution

${({NH}_{4})}_{2} {SO}_{4} + Ca {(OH)}_{2} \to {CaSO}_{4} . 2 H_{2} O + 2 {NH}_{3}$

1584 g gypsum ( M =172) 24mole

=12 mol = 12 mol

$\underset{952 g = 4 m o l}{N i C l_{2} .6 H_{2} O} + \underset{24 m o l}{6 N H_{3}} \to \underset{\begin{array}{l} (M = 232) \\ 4 m o l \end{array}}{[N i {(N H_{3})}_{6}]} C l_{2} + 6 H_{2} O$

Total mass $= 12 \times 172 + 4 \times 232 = 2992 g$

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