Correct Answer - `(1.04 xx 10^(4))`
Mass of `Fe_(2)O_(3) = 0.552g`
millimol of `Fe_(2)O_(3) = (0.552)/(160) xx 100 = 3.45`
During treatment with `Zn`-dust, all `Fe^(3+)` is reduced to `Fe^(2+)`, hence
millimol of `Fe^(2+)` is oxidised to `Fe^(3+)`, liberating one electron per `Fe^(2+)` ion. Therefore, total electrons taken up by oxidant.
`=1.725 xx 10^(-3) xx 6.023 xx 10^(23) = 1.04 xx 10^(21)`