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In the following change, $3 Fe + 4 H_{2} O ⟶ {Fe}_{3} O_{4} + 4 H_{2}$ If the atomic weight of iron is 56, then its equivalent weight will be :

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answer is B.

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Detailed Solution

$3 Fe + 4 H_{2} O ⟶ {Fe}_{3} O_{4} + 8 H^{+} + 8 e^{-}$

V.F. of $Fe = \frac{8}{3}$

$\Rightarrow E_{Fe} = \frac{Atomic mass}{V \cdot F .} = \frac{56}{8} = 21$

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