Question

For the following reaction:
$M^{x+}+Mn{O}_4^{\ominus }⟶M{O}_3^{\ominus }+{Mn}^{2+}+\left(\frac{1}{2}\right)O_2$
If $1$ mol of $Mn{O}_4^{\ominus }$ oxidises $1.67$ mol of $M^{x+}$ to $M{O}_3^{\ominus }$, then the value of $x$ in the reaction is :

• A. $5$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (C)

$2$

Given that, $Mn{O}_4^{-}$ oxidizes $M^{x+}$ to $M{O}_3^{-}$.
The reactions are given below:
$5e^{-}+Mn{O}_4^{\ominus }⟶{Mn}^{2+}$
$M^{x+}⟶M{O}_3^{\ominus }+(5-x)$
Equivalent of $Mn{O}_4^{\ominus }=$ Equivalent of $M^{x+}$
$1$ mol $\times 5=1.67$ mol $\times (5-x)$
$\therefore x=2$