A mixture of  ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$  and  ${\text{K}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ required $\text{0.2\hspace{0.17em}N,}$   25 mL  ${\text{KMnO}}_4$ solution for complete oxidation same mixture needs  $0.2\text{\hspace{0.17em}M}$ , 20 mL NaOH solution for its complete neutralization. Calculate mole percentage of  ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ in the given mixture.

Let mole of  ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ and  ${\text{K}}_2{\text{C}}_2{\text{O}}_4$ are x and y respectively.
Number of equivalent of   ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ + Number of equivalent of   ${\text{K}}_2{\text{C}}_2{\text{O}}_4$= Number of equivalent of${\text{KMnO}}_{\text{4}}$   

$$\begin{gathered} 2\times x+2y=0.2\times 25 \\ x+y=2.5 \end{gathered}$$

No. of equivalent of  ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}=$  Number of equivalent of $\text{NaOH}$

$$\begin{gathered} 2x=0.2\times 20=4 \\ x=2 y=0.5 \end{gathered}$$

Mole percentage of    ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}=\frac{2}{2.5}$