Solution

$$5\overline { e } +Mn{ O }_{ 4 }^{ -2 }\longrightarrow M{ n }^{ +2 }\times 2\\ +\quad \quad \quad { 5 }^{ -2 }+2\overline { e } \longrightarrow S\times 2\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ 2Mn{ O }_{ 4 }^{ - }+5{ S }^{ -2 }\longrightarrow 2Mn^{ +2 }+5S$$
Divide reaction by $$5.$$
$$\cfrac { 2 }{ 5 } Mn{ O }_{ 4 }^{ - }+{ 5 }^{ -2 }\longrightarrow 2Mn^{ +2 }+5S$$
Hence $$1$$ mole of sulphide ions will require $$\cfrac { 2 }{ 5 } $$ moles of $$Mn{ O }_{ 4 }^{ - }$$ ions.