$(II)\,\mathop {BrO_3^{- }}\limits^{ + 5}  + \mathop {AsO_2^ - }\limits^{ + 3}  + {H_2}O \to \mathop {B{r^ - }}\limits^{ - 1}  + \mathop {AsO_4^{3 - }}\limits^{ + 5}  + {H^ + }$

In reaction $(II)$

gm. eq. of $BrO_3^-=$ gm. eq. of $AsO_2^-$

$n_{BrO_3^-} \times 6=n_{AsO_2^-} \times 2$

$ = \frac{{12.5}}{{1000}} \times \frac{1}{{25}} \times 2 = {10^{ - 3}}$

${n_{BrO_3^ - }} = \frac{{{{10}^{ - 3}}}}{6}$

In reaction $(I)$

moles of $BrO_3^-$ consumed

$ = \frac{{70}}{{1000}} \times \frac{1}{{60}} - \frac{{{{10}^{ - 3}}}}{6} = {10^{ - 3}}$

gm eq. of $SeO_3^{2-} = gm. eq. of BrO_3^-$

$n_{SeO_3^{2-}} \times 2 = 10^{-3} \times 5$ ;

$n_{SeO_3^{2-}} = 2.5 \times 10^{-3}$