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A 3.00 g sample containing Fe3O4 , Fe2O3 and an inert impure substance, is treated with excess of KI solution in presence of dilute H2SO4. The entire iron is converted into Fe2+ along with the liberation of iodine. The resulting solution is dilute to 100 mL . A 20 mL of the diluted solution requires 11.0 mL of 0.5 M Na2S2O3 solution to reduce the iodine present. A 50 mL fo the dilute solution, after complete extraction of the iodine required 12.80 mL of 0.25 M KMnO4 solution in dilute H2SO4 medium for the oxidation of Fe2+. Calculate the percentage of Fe2O3 and Fe3O4 in the original sample.

A 3.00 g sample containing Fe3O4 , Fe2O3 and an inert impure substance, is treated with excess of KI solution in presence of dilute H2SO4. The entire iron is converted into Fe2+ along with the liberation of iodine. The resulting solution is dilute to 100 mL . A 20 mL of the diluted solution requires 11.0 mL of 0.5 M Na2S2O3 solution to reduce the iodine present. A 50 mL fo the dilute solution, after complete extraction of the iodine required 12.80 mL of 0.25 M KMnO4 solution in dilute H2SO4 medium for the oxidation of Fe2+. Calculate the percentage of Fe2O3 and Fe3O4 in the original sample.

Grade:10

1 Answers

Deepak Patra
askIITians Faculty 471 Points
9 years ago
Sol. Let the original sample contain x millimol of Fe3O4 and y millimol of Fe2O3. In the first phase of reaction. Fe3O4 + I- → 3Fe2+ + I2 (n-factor of Fe3O4 = 2) Fe2O3 + I- → 2Fe2+ + I2 (n-factor of Fe2O3 = 2) ⇒ Meq of I2 formed = Meq (Fe3O4 + Fe2O3) = Meq of hypo required ⇒ 2x + 2 y = 11 × 0.5 = 27.5 ….(i) Now, total millimol of Fe2+ formed = 3x + 2 y. in the reaction Fe2+ + MnO_4^- + H+ → Fe3+ + Mn2+ n-factor of Fe2+ = 1 ⇒ Meq of MnO_4^- = Meq of Fe2+ ⇒ 3x + 2 y = 12.8 × 0.25 × 5 = 27.5 ….(ii) Solving Eqs. (i) and (ii), we get X = 4.5 and y = 9.25 ⇒ Mass of Fe3O4 = 4.5/1000 × 232 = 1.044 g % mass of Fe3O4 = 1.044/3 × 100 = 34.80% Mass of Fe2O3 = 9.25/1000 × 160 = 1.48 g % mass of Fe2O3 = 1.48/3 × 100

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