Calculate the number of moles of `Cr_2O_7^(2-)` required to oxidise 1 " mol of "`Fe(HC_2O_4)` in acidic medium. How many moles of NaOH are required to react with 1 " mol of "`Fe(HC_2O_4)`?

Explanation

(a). It is a redox reaction:
`6e^(-)+Cr_2O_7^(2-)to2Cr^(3+)(n=6)`
`[{:(Fe^(2+)toFe^(3+)+e^(-)),(2(HC_2O_4^(ɵ))to4CO_2+4e^(2-)):}](n=5)`
`Cr_2O_7^(2-)-=Fe(HC_2O_4)_2`
`1Eq-=1Eq`
`(1)/(6)mol=(1)/(5)mol`
`1 " mol of "Fe(HC_2O_4)_2+(5)/(6)" mol of "Cr_2O_7^(2-)`
(b). It is a acid base reaction
`NaOH(n=1`, one `overset(ɵ)(O)H` ion), `(HC_2O_4^(ɵ))_2(n=2,2H^(o+)` ions)
`Fe^(2+)` does not react with base (NaOH)
`NaOH-=(HC_2O_4)_2`
`1Eq-=1Eq`
`(1)/(2)mol-=(1)/(2)mol`
`2 " mol of "NaOH-=1 " mol of "Fe(HC_2O_4)_2`