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 value's of $x,y$ and $z$ in the reaction are respectively:

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

Answer 1

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

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

$C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}$-----------(2)

Multiplying Equation (1) by (2) and (2) by 5, and adding them

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

$\therefore$, $x=2y=5z=16$