(a) $N_{2} - - - + 3 H_{2} \to 2 N H_{3} - ----- -$
(b) $C a (O H)_{2} + H_{3} P O_{4} - ------ - \to C a H P O_{4}$
(c) $Z n - -- - + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$
(d) $2 F e S O_{4} - ------- - + H_{2} S O_{4} + O \to F e (S O_{4})_{3} + H_{2} O$
The equivalent weights of the underlined (from top to bottom) based on the above reactions are $\frac{M}{6}$ , $\frac{M}{3}$ , $\frac{M}{2}$ , $\frac{M}{2}$ , $M$ respectively.
Taking molecular weight $= M$

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Solution

(a) $N_{2} = 6 H$
$∴ 1 H = \frac{N_{2}}{6} = \frac{28}{6} = \frac{M}{6}$
$2 N H_{3} = 6 H$
$∴ 1 H = \frac{2 N H_{3}}{6} = \frac{N H_{3}}{3} = \frac{17}{3} = \frac{M}{3}$
(b) $H_{3} P O_{4} = 2 O H^{-}$
$∴ 1 O H =^{-} = \frac{H_{3} P O_{4}}{2} = \frac{M}{2}$
(c) $Z n = 2 H^{+}$
$∴$ Equivalent weight of $Z n = \frac{M}{2}$
(d) $2 F e S O_{4} = 1 (O) = 16 g = 2 \times 8 g$
$∴ 8 g (O) = 1 F e S O_{4}$
$∴$ Equivalent weight of $F e S O_{4} = M$

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