Question

1 mol of ferric oxalate is oxidized by x mol of $Mn{O}_4^{-}$ and also 1 mol of ferrous oxalate is oxidised by y mol of $Mn{O}_4^{-}$ in acidic medium. The ration $\frac{x}{y}$ is:

• A. 2 : 1
• B. 1 : 2
• C. 3 : 1
• D. 1 : 3

Answer 1

The correct option is A 2 : 1

$F{e}_2\left(204\right)3+Mn{O}_4^{-}\to M{n}^{2+}+C{o}_2$
$1\times 6=x\times 5$
$x=\frac{6}{5}$
$Mn{O}_4^{-}+Fe{C}_2{O}_4\to M{n}^{2+}+F{e}^{3+}+C{o}_2$
4-factor = 2 + 1 = 3
$y\times 5=1\times 3$
$y=\frac{3}{5}$
$\frac{x}{y}=\frac{6}{5}/\frac{3}{5}=2:1$