Question

A mixture of $FeO$ and $Fe_2O_3$ is completely reacted with $100\, mL$ of $0.25 \,M$ acidified $KMnO_4$ solution. The resultant solution was then treated with $Zn$ dust which converted $Fe^{3+}$ of the solution to $Fe^{2+}$. The $Fe^{2+}$ required $1000 \,mL$ of $0.10 \,M \,K_2Cr_2O_7$ solution. Find out the weight percentage of $Fe_2O_3$ in the mixture.

Answer 1

$m$-eq. of $FeO = m$-eq. of $KM nO_4$
$ = 0.25 \times 5 \times 100$
$m$-mole of $FeO (n = 1) = \frac{0.25 \times 100 \times 5}{1} = 125$
Total $m$-eq. or $m$-mole. of $Fe^{2+}$
$= 1000 \times 0.1 \times 6 = 600$
(from $FeO$ and $Fe_2O_3$ after reaction with $Zn$ dust)
$m$-mole of $Fe^{2+}$ from $Fe_2O_3 = 600 -125 = 475$
$m$ mole of $FeO_3 = \frac{475}{2}$
Mass of $FeO = \frac{125 \times (56 + 16)}{1000} g = 9\,g$
Mass of $Fe_2O_3 = \frac{475}{2} \times \frac{160}{1000} = 38\,g$
$\% Fe_2O_3 = \frac{38}{38 + 9} \times 100 = 80.85 \%$