Question

Consider the following reaction:
$M^{x+}+Mn{O}_4^{\ominus }\to M{O}_3^{\ominus }+M{n}^{2+}$
If $1mol$ of $Mn{O}_4^{\ominus }$ oxidises $1.67mol$ of $M^{x+}$ to $M{O}_3^{\ominus }$, then the value of $x$ in the reaction is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (A)

$M^{x+}+Mn{O}_4^{\ominus }⟶M{O}_3{}^{\ominus }+M{n}^{2+}$
$Mn{O}_4{}^{\ominus }+8H^{\oplus }+5e^{-}⟶M{n}^{2+}+4H_{2}O$
$M^{x+}+3H_{2}O⟶M{O}_3{}^{\ominus }+6H^{\oplus }+(5-x)e^{-}$
$mEq$ of $Mn{O}_4{}^{2-}\equiv mEq$ of $M^{x+}$
$1\times 5\equiv [1.67\times (5-x)\Rightarrow 5-x=3\Rightarrow x=2$