5.5 g of a mixutre of `FeSO_4.7H_2O` and `Fe_2(SO_4)_3.9H_2O` requires 5.4 " mL of " `0.1 N KMnO_4` solution for complete oxidation. Calculate the number of gram moles of hydrated ferric sulphate in the mixture.

Explanation

Redox changes are:
`5e+Mn^(7+)rarrMn^(2+)`
`Fe^(2+)rarrFe^(3+)+1e`
It is no be noted here that only `FeSO_(4).7H_(2)O` will react with `KMnO_(4)` to bring in redox change.
`because` Meq.of `FeSO_(4).7H_(2)O= "Meq.of" KMnO_(4)`
`(w)/(E )xx1000=5.4xx0.1`
`therefore (w)/((278)/(1))xx1000=0.54`
`therefore w=0.150g`
`therefore` Weight of `Fe_(2)(SO_(4))_(3).9H_(2)O=5.5-0.150g`
`=5.350 g`
`therefore` Mole of `Fe_(2)(SO_(4))_(3).9H_(2)O=(5.350)/(562)`
`=9.5xx10^(-3)"mole"`
`( because "M.wt." of `Fe_(2)(SO_(4))_(3).9H_(2)O=562)`