A 3.33 gram sample of iron ore is transformed to a solution ofiron (II) sulfate

Question

A 3.33 gram sample of iron ore is transformed to a solution ofiron (II) sulfate and this solution is titrated with 0.150 Mpotassium dichromate. It it required 41.4 mL of potassiumdichromate solution to titrate the iron (II) solution, what is thepercentage of iron in the ore? Fe]2+ + CrO7]2- >>> Fe]3+ +Cr]3+      (unbalanced) A 3.33 gram sample of iron ore is transformed to a solution ofiron (II) sulfate and this solution is titrated with 0.150 Mpotassium dichromate. It it required 41.4 mL of potassiumdichromate solution to titrate the iron (II) solution, what is thepercentage of iron in the ore? Fe]2+ + CrO7]2- >>> Fe]3+ +Cr]3+      (unbalanced)

Explanation / Answer

Fe (II) + 2Cr (VI) ...........> Fe(III) + 2Cr (III) from the above equation it is clear that each Cr (VI) ion canoxidize three Fe(II) ions . Hence moles of Fe(II)oxidizied is equals to 3/2 moles of Cr(VI). moles of Fe(II) = (3/2)* (0.150M*41.4mL)                        = 9.315 m mol mass of Fe(II) = 9.315 m mol * 55.845 g/ mol                       = 520.196 mg                       = 0.521 g so percentage of iron in the ore = (0.521g / 3.33g)*100                                                = 15. 62%

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