Find the number of moles of `KMnO_(4)` needed one mole `Cu_(2)S` in acidic medium.
The reaction is `KMnO_(4) + Cu_(2)S rarr M n^(2+) + Cu^(2+) + SO_(2)`

Explanation

From law of equivalence,
equivalents of `Cu_(2)S` = equivalents of `KMnO_(4)`
moles of `Cu_(2)Sxx` v.f. = moles of `KMnO_(4)xx` v.f.
`1xx8="moles of" KMnO_(4)xx5 rArr "moles of " KMnO_(4) = 8//5"`
`(.:. "v.f. of "Cu_(2)S=2(2-1)+1(4-(-2)))=8 " and v.f. of " KMnO_(4) = 1(7-2)=5)`