\(\mathrm{n}_{\mathrm{CO}_2}=\frac{22}{44}=0.5 \mathrm{moles}\)
So moles of \(\mathrm{CH}_4\) required \(=0.5\) moles i.e. \(50 \times 10^{-2} \mathrm{~mole}\)
x=\(50\)
$I.\,SeO_3^{2 - } + BrO_3^ - + {H^ + } \to SeO_4^{2 - } + B{r_2} + {H_2}O$
$II.\,BrO_3^ - + AsO_2^ - + {H_2}O \to B{r^ - } + AsO_4^{3 - } + {H^ + }$