Question

The weight of `KMnO_(4)` required to completely oxidise 0.25 moles of `FeSO_(4)` in acid medium is...........(molecular weight of `KMnO_(4) = 158`)
A. 5.8
B. 1.5
C. 7.9
D. 0.79

Answer 1

Correct Answer - C
`2KMnO_(4) +3H_(2)SO_(4) +10 FeSO_(4) rarr K_(2)SO_(4) +2MnSO_(4) +5Fe_(2)(SO_(4))_(3) +3H_(2)O`
`2 xx 158 gm` of `KMnO_(4)`
`rarr 10` moles of `FeSO_(4)`?
`-0.25` moles `FeSO_(4)`