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Question

# A 3.0 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 to Fe2+ along with the liberation of iodine. The resulting solution is diluted to 100 mL. A 20 mL of the dilute solution requires 11.0 mL of 0.5 M Na2S2O3 solution to reduce the iodine present. A 50 mL of the diluted solution after complete extraction of iodine requires 12.8 mL of 0.25 M KMnO4 solution in dilute H2SO4 medium for oxidation of Fe2+. Calculate the percentage of Fe2O3 and Fe3O4 in the original sample.

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Solution

## Fe3O4 is an equimolar mixture of Fe2O3 and FeO. Thus, the sample contains Fe2O3,FeO and impurities. The amount of iodine libeated depends on the amount of Fe2O3 and the entire iron is converted into Fe2+.Fe3O4+2KI+H2SO4→3FeO+H2O+K2SO4+I2Fe2O3+KI+H2SO4→2FeO+H2O+K2SO4+I25×11.0 mL of 0.5 M Na2S2O3≡55.0 mL of 0.5 N Na2S2O3soln.≡55.0 mL of 0.5 N I2soln.≡55.0 mL of 0.5 N Fe2O3soln.=27.5×10−3 equivalent Fe3O4 soln.=13.75×10−3molesFe2O32×12.8 mL of 0.25 M KMnO4soln.≡25.6 mL of 1.25 N KMnO4soln.≡25.6 mL of 1.25 N FeO soln.=32.0×10−3 equivalent FeO=32.0×10−3moles FeOMoles of FeO in Fe2O4=0.032−0.0275=0.0045Mass of Fe3O4=0.0045×232=1.044 gMoles of Fe2O3 existing separately=0.01375−0.0045=0.00925% Fe3O4=1.0443×100=34.8% Fe2O3=1.483×100=49.33.

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