$\mathrm{n}_{\mathrm{CO}_2}=\frac{22}{44}=0.5 \mathrm{moles}$
So moles of $\mathrm{CH}_4$ required $=0.5$ moles i.e. $50 \times 10^{-2} \mathrm{~mole}$
x=$50$
$2MnO_4^ - + 5{C_2}O_4^ - + 16{H^ + } \to 2M{n^{ + + }} + 10C{O_2} + 8{H_2}O$
અહી $20\, mL$ of $0.1\, M\, KMnO_4$ એ કોના બરાબર હશે