H _ 2 S ( 5 moles) reacts completely with acidified aqueous...

${KMnO}_{4}\overset{{H}^{+}}{⟶}{Mn}^{2 +}$

${S}^{- 2} ⟶ \overset{0}{S}$

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

n-factor of ${KMnO}_{4} = 5$

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

Moles \times n - factor = numberofequivalents$</p><p>\left({n}_{{KMnO}_{4}} \times 5\right) = ( 5 \times 2{)}_{{H}_{2}S}$

$∴ {n}_{{KMnO}_{4}} = 2$

$∴ 2{KMnO}_{4} + 3{H}_{2}{SO}_{4} + 5{H}_{2}S \to {K}_{2}{SO}_{4} + 2{MnSO}_{4} + 5S + 8{H}_{2}O$

Number of moles of water produced $= 8$

Number of moles of electrons involved $= 10$

∴ x = 8, y = 10 ⇒ ( x + y ) = 18