Equivalent weight of $F e S_{2}$ [mol.wt = M] in the following reaction is?
$F e S_{2} + O_{2} \to F e_{2} O_{3} + S O_{2}$

\frac{M}{2}

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\frac{M}{4}

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\frac{M}{11}

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\frac{M}{14}

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Solution

The correct option is C $\frac{M}{11}$
total electron involve in $= 24 - 2 = 22$
Number of electron involve in reduction by one molecule
$of {F e S}_{2} = \frac{22}{2} = 11$
$n -$ factor of $F e s_{2} = 11$
Equivalent $w t = \frac{molecular wt}{n - factor}$
$= \frac{m}{11}$
Correet dnswer - $C - -$

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