Chemistry - Redox Reactions Question with Solution | TestHub

As per as the balanced equation given below
$3{\mathrm{S}}_2{\mathrm{O}}_3^{2-}+8\mathrm{M}\mathrm{n}{\mathrm{O}}_4^{-}+{\mathrm{H}}_2\mathrm{O}\to 8\mathrm{M}\mathrm{n}{\mathrm{O}}_2+6\mathrm{S}{\mathrm{O}}_4^{2-}+2\mathrm{O}{\mathrm{H}}^{-}$
$3$moles of$\mathrm{N}{\mathrm{a}}_2{\mathrm{S}}_2{\mathrm{O}}_3=8$moles of$\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4$
$3\times 248$g of$\mathrm{N}{\mathrm{a}}_2{\mathrm{S}}_2{\mathrm{O}}_3=8\times 158$g of$\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4$
0.158 g of$\mathrm{N}{\mathrm{a}}_2{\mathrm{S}}_2{\mathrm{O}}_3=\frac{8\times 158\times 0.158}{3\times 248}$
$=0.27$g$\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4$
Volume$=\frac{\mathrm{W}\mathrm{t}.\times 1000}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}\times \mathrm{M}\mathrm{o}\mathrm{l}.\mathrm{W}\mathrm{t}.}$
$=\frac{0.27\times 1000}{0.1\times 158}=\mathrm{17\; ml}$.