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InterviewSolution
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A 6.00g sample contained Fe_(3) O_(4), Fe_(2) O_(3) and inert materials. It was treated with an excess of aqueous Kl in acidic medium which reduced all the iron to Fe^(2+). The resulting solution was dilution to 50ml.and a 10ml.sample of it was taken. The liberated iodine reacts with 5.5ml. of 1M Na_(2) S_(2) O_(3) solution, yielding S_(4) O_(6)^(2-). The iodine from another 25ml. sample was extracted, after which the Fe^(2+) was titrated with 3.2 ml. of 1M MnO_(4)^(-) in H_(2)SO_(4) solution. Calculate the percentage of Fe_(3) O_(4) and Fe_(2)O_(3) in the original mixture. |
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Answer» Solution :Equivalents of `Na_(2)S_(2) O_(3)` reacting with the liberated `I_(2) = ( 5.5 xx 1)/( 1000) = 5.5 xx 10^(-3)` `:.` Equivalent of `I_(2)` in 10ml `= 5.5 xx 10^(-3)` Equivalent of `Fe^(3+)` in 50ml = `5.5 xx 10^(-3) xx 5 = 0.0275`[ lt is `Fe^(3+)` that reacts with KL to give `I_(2) `] Total moles of `Fe_(2) O_(3) = ( 0.0275)/( 2) = .01375` ( free `+` combined ) Equivalengt of `MnO_(4)^(-) = ( 1 xx 5 xx 3.2 )/( 1000) = 0.016` `:.` Eq of `Fe^(2+)` in 25 ml= 0.016 Eq of `Fe^(2+)` in 50ml = `0.016 xx 2 = 0.0 32 ` Eq. of FeO `= 0.032 - 0.0275 = 4.5 xx 10^(-3)` Moles of `FeO = 4.5 xx 10^(-3)` `:.` Moles of `Fe_(3) O_(4) = 4.5 xx 10^(-3)` Mass of `Fe_(3) O_(4) = 4.5 xx 10^(-3) xx 232= 1.044g m`. Moles of `Fe_(2) O_(3) ` present FREELY `= 0.01375 - 4.5 xx 10^(-3) = 9.25 xx 10^(-3)` Mass of `Fe_(2) O_(3) ` present freely `= 9.25 xx 10^(-3) xx 160 = 1.48 g ` % of `Fe_(3) O_(4) = ( 1.044)/(6) xx 100 = 17.4 % ` %`Fe_(2) O_(3) = 24.67%` |
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