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5.5 g of a mixutre of FeSO4.7H2O and Fe2...

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.

Text Solution

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Only `FeSO_(4). 7H_(2)O` will be oxidised by `KMnO_(4)`.
Mol. Mass of `FeSO_(4). 7H_(2)O=278`
As the conversion involves one electron,
`Fe^(2+) to Fe^(3+) + e^(-)`,
The eq. mass of `FeSO_(4). 7H_(2)O=(278)/(1)=278`
5.4 mL of 0.1N `KMnO_(4)`
`-=5.4 mL " of " 0.1 N FeSO_(4). 7H_(2)O` solution
Amount of `FeSO_(4).7H_(2)O=(0.1xx278)/(1000)xx5.4=0.15g`
Amount of `Fe_(2)(SO_(4))_(3). 9H_(2)O=(5.5-0.15)=5.35g`
Mol. mass of `Fe_(2)(SO_(4))_(3).9H_(2)O=("Mass")/("Mol. mass")`
`=(5.35)/(562)=0.00952`
`=9.52xx10^(-3)`
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