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

Consider the following redox change
`Fe^(2+) to Fe^(3+) 1e " "5e+Mn^(7+ to Mn^(2)`
In this redox process, only `Fe^(2+)` are oxidised while `Fe^(3+)` ions remain unaffected.
Now, meq. Of `FeSO_(4). 7H_(2)O="meq. Of KMnO"_(4) w/E xx 1000=(w)/(278/1)xx 1000=5.4 xx 0.1`
`(E=M/1=278/1)`
w=0.150g [=mass of Fe(II) and Fe(III) sulphates]
Mass of `Fe_(2)(SO_(4)_(3)). 9H_(2)O=5.5 -0.150g=5.350g`
`n_(Fe_(2)(SO_(4))_(3). 8H_(2)O)=(5.350)/(562)=9.5 xx 10^(-3)" mole"`