One litre of acidified `KMnO_(4)` solution containing 15.8g `KMnO_(4)` is decolourized by passing sufficient `SO_(2)` . If `SO_(2)` is produced by `FeS_(2)` , what is the amount of `FeS_(2)` required to give desired `SO_(2)`?

Explanation

v.f. of `KMnO_(4)=5 ` & v.f. of `SO_(2)=2`
Now, Eq. of `KMnO_(4)`=Eq. of `SO_(2)`
`(15.8)/(158//5)`=moles of `SO_(2)xx2`
so, moles of `SO_(2)=1//4`
Now, applying POAC on S, we get :
`2xx` mole of `FeS_(2)=1xx` moles of `SO_(2)`
so, moles of `FeS_(2)=(1)/(4)xx(1)/(2)=(1)/(8)`
so, weight of `FeS_(2)=(1)/(8)xx120=15g`.