Chemistry - Complete d-block Question with Solution | TestHub

${\mathrm{KMnO}}_4\stackrel{{\mathrm{H}}^{+}}{⟶}{\mathrm{Mn}}^{2+}$

${\mathrm{S}}^{-2}⟶\stackrel{0}{ \mathrm{S}}$

The number of electrons involved per molecule is called n-factor.

n-factor of ${\mathrm{KMnO}}_4=5$

n-factor of ${\mathrm{S}}^{-2}\left({\mathrm{H}}_2 \mathrm{S}\right)=2$

$\mathrm{Moles}\times \mathrm{n}-\mathrm{factor} = \mathrm{number} \mathrm{of} \mathrm{equivalents}$ 

$\left({\mathrm{n}}_{{\mathrm{KMnO}}_4}\times 5\right)=(5\times 2{)}_{{\mathrm{H}}_2 \mathrm{S}}$

$\therefore {\mathrm{n}}_{{\mathrm{KMnO}}_4}=2$

$\therefore 2{\mathrm{KMnO}}_4+3{\mathrm{H}}_2{\mathrm{SO}}_4+5{\mathrm{H}}_2 \mathrm{S}\to {\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{MnSO}}_4+5 \mathrm{S}+8{\mathrm{H}}_2\mathrm{O}$

Number of moles of water produced $= 8$

Number of moles of electrons involved $=10$

$\therefore \mathrm{x}=8,\mathrm{y}=10\Rightarrow (\mathrm{x}+\mathrm{y})=18$