Question

Consider the following reaction:

$xMn{O}_4^{-}+y{C}_2{O}_4^{2-}+z{H}^{+}\to xM{n}^{2+}+2yC{O}_2+\frac{z}{2}{H}_{2}O$

The values of $x,y$ and $z$ in the reaction are, respectively :

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

Answer 1

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

On balancing the charge on both sides of reaction, we get

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

Hence, $x=2,y=5,z=16$.