If the mass of $KH{C}_2{O}_4$ required to reduce 100 ml of 0.02 M $KMn{O}_4$  in acidic medium (to Mn⁺²)   is ‘x’ gram and to neutralize 100 ml of 0.05 M Ca ${\left(OH\right)}_2$ is ‘y’ gram, then calculate the value of $\frac{x}{y}$  

$5{\mathrm{HC}}_2{\mathrm{O}}_4^{-}+2{\mathrm{MnO}}_4^{-}\to 2{\mathrm{Mn}}^{+2}+10{\mathrm{CO}}_2$

The number of moles of ${\mathrm{MnO}}_4^{-}$ =0.1×0.02=0.002 mol
This is equal to 0.005 mol of ${\mathrm{HC}}_2{\mathrm{O}}_4^{-}$ ion. This corresponds to x g.

The number of moles of Ca(OH)_2​ is 0.1×0.05=0.005 mol

They corresponds to 0.01 mol of hydrogen ions and 0.01 mol of ${\mathrm{HC}}_2{\mathrm{O}}_4^{-}$.
This is equal to y g.

$$\begin{gathered} \frac{\mathrm{x}}{\mathrm{y}}=\frac{0.005}{0.01}=\frac{1}{2} \\ \mathrm{y}=2\mathrm{x} \end{gathered}$$