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A 4.0 g sample contained Fe_2O_3,Fe_3O_4, and inert material. It was treated with an excess of aq KI solution in acidic medium, which reduced all iron to Fe^(2+) ions. The resulting solution was diluted to 50 mL and a 10 mL sample of it was taken the iodine liberated in the small sample was titrated with 12.0 " mL of " 0.5 M Na_2S_2O_3 solution. The iodine from another 25 mL was extracted, after which the Fe^(2+) ions were titrated with 16 " mL of " 0.25 M MnO_4^(ɵ) ions in H_2SO_4 solution. Calculate the mass of two oxides in the original mixture.

Answer» <html><body><p></p>Solution :Let `Fe_2O_3-=xmmol` <br/> `Fe_3O_4-=ymmol-=FeO+Fe_2O_3` <br/> `Fe^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>+)=2(x+y)mmol and Fe^(2+)=y m mol` <br/> `2Fe^(3+)+2I^(ɵ)toI_2+2Fe^(2+)` <br/> `2m" mol of "Fe^(3+)-=1 mmol I_2` <br/> `mmol I_2-=2m" mol of "S_2O_3^(2-)` (in 20 mL sample) <br/> `(x+y)(10)/(50)=(1)/(2)xx12xx0.5` <br/> `impliesx+y=15` <br/> <a href="https://interviewquestions.tuteehub.com/tag/note-1125278" style="font-weight:bold;" target="_blank" title="Click to know more about NOTE">NOTE</a> that now m" mol of "`Fe^(2+)` in 50 mL sample is <br/> `2(x+y)+y=2x+3y` <br/> Also, <br/> `5Fe^(2+)+MnO_4^(ɵ)to5Fe^(3+)<a href="https://interviewquestions.tuteehub.com/tag/mn-548487" style="font-weight:bold;" target="_blank" title="Click to know more about MN">MN</a>^(2+)` <br/> `5m" mol of "Fe^(2+)-=1 m" mol of "MnO_4^(ɵ)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a>(25)/(50)(2x+3y)=5(16xx0.25)` <br/> `implies2x+3y=40` <br/> Solving, we get `x=5 and y=10` <br/> Mass of `Fe_2O_3=(5)/(1000)xx160=0.8g` <br/> Mass of `Fe_2O_4=(10)/(1000)xx232=2.32g`</body></html>


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