Please help with steps to solve. The amount of iron in a sample can be determine

Question

Please help with steps to solve. The amount of iron in a sample can be determined by a titration using dichromate ion in acidic solution. The iron is first converted to Fe^2+, which is then titrated by the dichromate ion. The reaction is: 6Fe^2+(aq) + Cr_20_7^2+(aq) + 14 H^+ rightarrow 6 Fe^3+(aq) + 2 Cr^3+(aq) + 7 H_20 How many grams of iron are present in a sample if it required 42.7 ml of 0.0180 M Na_2Cr_2O_7(aq) solution for the titration described by the redox equation above? 0.043 g 0.258 g 3-61 g 7.07 g 7.15 g

Explanation / Answer

Balanced redox reaction is,

6Fe2+(aq.) + Cr2O72-(aq.) + 14 H+(aq.) ------- > 6Fe3+(aq.) + 2Cr3+(aq.) + 7H2O.

Oxidation half: 6Fe2+ ------------ > 6Fe3+ + 6e- ………….(1 e- per Fe transeffred)

Reduction half : Cr2O72- + 6 e- -------------- > 2Cr3+

In this titration the analyte is oxidized from Fe2+ to Fe3+, and the titrant is reduced from Cr6+ to Cr3+.

Oxidation of Fe2+ requires 1 e-. But to reduce 2 Cr6+ in Cr2O72- will require total of (2x3 = 6e-).

For conservation of electrons for redox reaction therefor requires that,

Moles of Fe2+ = 6 x moles of Cr2O72- …………..(1)

For Na2Cr2O7 we have,

Volume = 42.7 mL = 42.7 x 10-3 L ..............(Don't forget to convert mL to L by multiplying 10-3)

Molarity = 0.0180 M

Hence, Moles of Cr2O72- = Volume x Molarity .......................(Molar equation MV used)

Moles of Cr2O72- = (42.7 x 10-3) x (0.0180)

Moles of Cr2O72- = 7.686 x 10-4 moles ………………..(2)

And Moles of Fe in sample = Mass of in analyte/ Atomic weight of Fe .........(usual formula to find number moles)

Moles of Fe = Mass of Fe2+ /56 ………….. (3)

Using eq.(2) and (3) in eq.(1)

Mass of Fe in analyte / 56 = 7.686 x 10-4

Mass of Fe in analyte = 56 x 7.686 x 10-4

Mass of Fe in analyte = 0.043 g

i.e. 0.043 g Fe present in the analyte.

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