Calculate the Equivalent weight of $N H_{3}$ in the change $N_{2} \to N H_{3}$ .

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Solution

$N_{2} \to N H_{3}$
$N \to O$ oxidation state $N \to + 3$ Oxidation state
$N^{+ 3} H_{3} \to N_{2}^{0}$
$1$ mole of $N \times 3 \to 0$
nf of $N H_{3} = 3 - 0 = 3$
Equivalent weight of $N H_{3} = \frac{M o l a r m a s s o f N H_{3}}{n f o f N H_{3}}$
Equivalent weight $= \frac{17}{3} g / m o l e$

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