The ammonia prepared by treating ammonium sulphate with NaOH is used $C u S O_{4} .5 H_{2} O$ to form a stable coordination compound. If 330g of pure ammonium sulphate and 499g of pure $C u S O_{4} .5 H_{2} O$ are used in preparation and $N a O H$ is used in excess, The moles of Copper – Ammonia coordination compound produced is Y. The Value of 4Y is equal to
(Atomic weight in $g m o l^{-} H = 1, N = 14, O = 16, S = 32, N a = 23, C u = 63.5$ )

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answer is 5.

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Detailed Solution

${(N H_{4})}_{2} S O_{4} + 2 N a O H \to 2 N H_{3} + N a_{2} S O_{4} + 2 H_{2} O n = 2.5$

$\underset{n = 2}{C u S O_{4}} .5 H_{2} O + \underset{\begin{array}{l} n = 5 \\ limiting \end{array}}{4 N H_{3}} \to [C u {(N H_{3})}_{4}] S O_{4} + 5 H_{2} O$

Hence, moles of $[C u {(N H_{3})}_{4}] S O_{4} = \frac{5}{4} = 1.25$

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