Question

The number of moles of $KMn{O}_4$ reduced by 1 mol of $KI$ in alkaline medium is :

• A. $1$
• B. $2$
• C. $5$
• D. $\frac{1}{5}$

Answer 1

The correct option is B $2$

In a basic medium,
$3e^{-}+Mn{O}_4^{\ominus }⟶Mn{O}_2$
$I^{\ominus }⟶I{O}_3^{\ominus }+6e^{-}$
Equivalent of $Mn{O}_4^{\ominus }=$ Equivalent of $I^{\ominus }$

$\frac{6}{3}$ mol of $Mn{O}_4^{\ominus }=1$ mol of $I^{\ominus }$
$\therefore 2$ mol of $Mn{O}_4^{\ominus }=1$ mol of $I^{\ominus }$