What is the equivalent weight of ${NH}_{3}$ in the given reaction?
$3 CuO + 2 {NH}_{3} \to 3 Cu + N_{2} + 3 H_{2} O$

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$\frac{17}{4}$

$\frac{17}{3}$

$\frac{17}{2}$

answer is D.

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Detailed Solution

$2 N^{3 -} \to {(N_{2})}^{o} + 6 e^{-} 2 mole of {NH}_{3} = 1 mole N_{2} Thus equivalents 2 \times n = 1 \times 6 n = 6 / 2 = 3 ∴ Eq . wt . = \frac{M}{3} = \frac{17}{3}$

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