Question

Coefficients of $Mn{O}_4^{2-},C_2{O}_4^{2-}$ and $H^{+}$ in the balanced reaction.
$Mn{O}_4^{-}+C_2{O}_4^{2-}+H^{+}\to M{n}^{2+}+C{O}_2+H_{2}O$

• A. $2,3$ and $8$
• B. $2,5$ and $8$
• C. $2,3$ and $16$
• D. $2,5$ and $16$

Answer 1

The correct option is D $2,5$ and $16$

$Mn{O}_4^{-}+C_2{O}_4^{2-}+H^{+}⟶M{n}^{2+}+C{O}_2+H_{2}O$

Now, here $Mn{O}_4^{-}$ is reduced to $M{n}^{2+}.$

$M{n}_4^{-}+8H^{+}+5e^{-}⟶M{n}^{2+}+4H_{2}O⟶\left(1\right)$

again. $C_2{O}_4^{2-}-2e^{-}⟶2C{O}_2⟶\left(2\right)$

Now, multiply $2$ with equation $\left(1\right)$ and $5$ with equation $\left(2\right)$, we will get,

$2Mn{O}_4^{-}+16H^{+}+5C_2{O}_4^{2-}⟶2M{n}^{2+}+8H_{2}O+10C{O}_2$

$\therefore$ Co-efficients of $Mn{O}_4^{-},C_2{O}_4^{2-}$ and $H^{+}$ in the balanced equation is $2$ , $5$ and $16$.