`0.804 g` sample of iron ore was dissolved in acid. Iron was oxidised to `+2` state and it requires `47.2 mL` of `0.112N KMnO_(4)` solution for titration, Calculate `%` of `Fe` and `Fe_(3)O_(4)` in ore.
Text Solution
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The titration involves the conversion of ferrous into ferric. `5Fe^(2+)+MnO_(4)^(-) +8H^(+) to 5Fe^(3+)+Mn^(2+)+4H_(2)O` `47.2 mL " of" 0.112 N KMnO_(4)-=47.2 mL " of" 0.112 N Fe^(2+)`ions `=(47.2xx0.112xx55.5)/(1000)=0.2934` Mass of iron =0.2934 g % of iron in the ore`=(0.2934)/(0.804)xx100=36.49` `underset((3xx55.5))(3Fe) to underset((3xx55.5+64))(Fe_(3)O_(4))` `3xx55.5` g of iron form 230.5 g of `Fe_(3)O_(4)`. 0.2934 g of iron will form `=(230.5)/(166.5)xx0.2934=0.406g` % of `Fe_(3)O_(4)` in the ore `= (0.406)/(0.804)xx100=50.5`
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