InterviewSolution
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InterviewSolution
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A 8. 0 g sample contained Fe_(3)O_(4) , Fe2O3 and inert materials . It was treated with an excess of aqueous Kl Solution in acidic medium,which reduced all the iron to Fe^(+2) ions. The resulting solution was diluted to 50.0 cm^(3) and a 10. 0 cm^(3) of it was taken. The liberated iodine in this solution required 7. 2 cm^(3) of 1. 0 M Na_(2) S_(2)O_(3) for reduction to iodide. The iodine from another 25. 0 cm^(3) sample was extracted , after which the Fe^(+2) ions was titrated against 1. 0 M MnO_(4)^(-)in acidic medium. The volume of KMnO_(4)solution used was found to be 4. 2 cm^(3) . Calculate the mass percentages of Fe_(3)O_(4)and of Fe_(2)O_(3) in the original mixture |
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Answer» Solution :This PROBLEM can be done by two methods. In the first method, we break up `Fe_(3) O_(4)` as an equimolar mixture of `FeO` and `Fe_(2) O_(3)` Method ( 1) `Fe_(3)O_(4) ` is` FeO. Fe_(2)O_(3)` Equivalents of `Na_(2)S_(2) O_(3) = ( 1 xx 7.2 ) /( 1000)= 7.2 xx 10^(-3)` Equivalents of `I_(2)` in 10cc `= 7.2 xx 10^(-3)` Equivalents of `I_(2)` in 50 CC `= 7.2 xx 10^(-3) xx 5= 3.6 xx 10^(-2)` Since equivalents of `I_(2)` is equal to that of KI which in turn is equal to the total equivalents of `Fe_(2) O_(3) ( Fe_(2) O_(3)` in the free state and `Fe_(3) O_(3)` COMBINED with `FeO)` `:.` equivalents of `KMnO_(4)` solution ` = ( 4.2 xx 1 xx 5)/( 1000) = 2.1 xx 10^(-2)` Since `KMnO_(4)` reacts with the total `Fe^(2+) ( Fe^(2+)` in FeO and `Fe^(2+)` that was produced by the action of Kl on `Fe_(2) O_(3))` `:.` equivalents of total `Fe^(2-)` in 50ml `= 2.1 xx 10^(-2) xx 2 = 4.2 xx 10^(-2)` Since equivalent of `Fe^(2+)` produced from `Fe_(2)O_(3)` is equal to that of equivalents of `Fe_(2) O_(3)` `:.` Equivalents of `FeO` `= 4.2 xx 10^(-2) - 3.6 xx 10^(-2) = 6 xx 10^(-3)` `:.` moles of `FeO = 6 xx 10^(-2)` moles of `Fe_(2)O_(3)` combined with FeO `= 6 xx 10^(-3)` total moles of `Fe_(2) O_(3) = ( 3.6 xx 10^(-2))/( 2)` ( because when `Fe_(2) O_(3) gt Fe^(2+) ` 'n' factor is 2 ) ` = 1.8 xx 10^(-2)` moles of `Fe_(2) O_(3)` in the free state `= 1.8 xx 10^(_2) - 6 xx 10^(-3) = 1.2 xx 10^(-2)` mass of `Fe_(3) O_(4) = 6 xx 10^(-3) xx 232 = 1.392g ` mass of `Fe_(3) O_(4) = 1.2 xx 10^(-2) xx 160 = 1.92 g ` percentage `Fe_(2) O_(4) = 17.4%` percentage of `Fe_(2) O_(3) = 23.75 %` Method2`:` Here we take `Fe_(3) O_(4)` as a single entity. Equivalents of `Na_(2) S_(2) O_(3) =( 7.2 xx 1)/( 1000) = 7.2 xx 10^(-3) ` Equivalents of `I_(2) ` in 50 cc `= 7.2 xx 10^(-3) xx 5 = 3.6 xx 10^(_2)` `:.` Equivalents of `Fe_(3) O_(4) +Fe_(2) O_(3) = 3.6 xx 10^(_2)` Let us assume that the moles of `Fe_(3) O_(4)` in xg and that of `Fe_(2) O_(3)` is y g . Since on reacting with KI both `Fe_(3) O_(4)` and `Fe_(2) O_(3)` give `Fe^(2+)` 'n' factor for both is two . `:. 2x + 2y = 3.6 xx 10^(-3) `........(1) Equivalent of `KMnO_(4)= ( 4.2 xx 1 xx 5 ) /( 1000) = 2.1 xx 10^(-2)` moles of `Fe^(2+)` in 50 mL `= 4.2 xx 10^(-2)` Since the moles of `Fe_(3) O_(4)` are x, moles of `Fe^(2+)` produced from `Fe_(3) O_(4)` will be 3x and that produced from `Fe_(2) O_(3) `will be 2y `:. 3x + 2y = 4.2 xx 10.2`.....(2) `( 2) - ( 1) ` gives `x = 6 xx 10^(-3)` `:. y = 1.2 xx 10^(-2)` Solving this percentage of `Fe_(3) O_(4)` is 17.4 % and `Fe_(2) O_(3)` is 23.75% |
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