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.

5.5 g of a mixutre of `FeSO_4.7H_2O` and `Fe_2(SO_4)_3.9H_2O` requires

1.	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.
Answer» 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)`

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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.

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