Question

Equivalent weight of $FeS _{2}$ (mol. wt. $=M$ ) in the following reaction is
$FeS _{2}+ O _{2} \rightarrow Fe _{2} O _{3}+ SO _{2}$

• A. $M / 2$
• B. $M / 4$
• C. $M / 11$
• D. $M / 14$

Answer 1

Answer: (C)

Calculate total change in oxidation number (O.N.) of one molecules of $FeS _{2}$

$FeS _{2}= Fe ^{2+}+2 S ^{-}$

$Fe ^{2+} \rightarrow Fe ^{3+}\left(\right.$ in $\left.\frac{1}{2} Fe _{2} O _{3}\right),$ Increase $= 1$

$2 S ^{-} \rightarrow 2 \overset{+4}{SO _{2}}, $ Increase $=10$

Total Increase $=11$

$\therefore $ Equivalent wt. = Molecular weight $/11$