How many mL of a 0.05 M `KMnO_(4)` solution are required to oxidise 2.0 g of `FeSO_(4)` in a dilute solution (acidic) ?

Explanation

`10FeSO_(4) + 2KMnO_(4) + 8H_(2)SO_(4) to K_(2)SO_(4) + 2MnSO_(4) + 5Fe_(2)(SO_(4))_(3) + 8H_(2)O`
`10 xx 151.8 " " 2 xx 158`
` 10 xx 151.8 "g of " FeSO_(4)` require `KMnO_(4) = 2 xx 158 g`
2 g of `FeSO_(4)` will require `KMnO_(4) = (2 xx 158 xx 2)/(10 xx 151 : 8) ` g
Suppose V mL solution (`0.05 M`) is required.
Amount of `KMnO_(4)` in this solution = `(158 xx 0.05)/1000 xx V`
Thus, `(158 xx 0.05 xx V)/1000 = ( 2 xx 158 xx 2)/(10 xx 151 .8) `
`V = 52.7 ` mL