The equivalent weight of FeS₂ in the following reaction is ${FeS}_{2} + O_{2} \to {Fe}^{+ 3} + {SO}_{2}$

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$\frac{Mol. Wt}{1}$

$\frac{Mol. Wt}{9}$

$\frac{Mol. Wt}{7}$

$\frac{Mol. Wt}{11}$

answer is C.

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Detailed Solution

${FeS}_{2} ⟶ {Fe}^{2 +} and S_{2}^{2 -}$
i.e., Sulphur is in -1 state
$∴ For 1 mole of {FeS}_{2}, ΔO \cdot N . = + 11$
$\Rightarrow n factor = 11$
$ΔO . N . = change in oxidation number .$

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