Question

A mixture of $FeO$ and $F{e}_2{O}_3$ is completely reacted with $100mL$ of $0.25M$ acidified $KMn{O}_4$ solution. The resultant solution was then treated with $Zn$ dust which converted $F{e}^{3+}$ of the solution to $F{e}^{2+}$. The $F{e}^{2+}$ required $1000mL$ of $0.10M{K}_{2}C{r}_2{O}_7$ solution. Find out the weight percentage of $F{e}_2{O}_3$ in the mixture.

Answer 1

Answer: 80.85

$m$-eq. of $FeO=m$-eq. of $KMn{O}_4$
$=0.25\times 5\times 100$
$m$-mole of $FeO(n=1)=\frac{0.25\times 100\times 5}{1}=125$
Total $m$-eq. or $m$-mole. of $F{e}^{2+}$
$=1000\times 0.1\times 6=600$
(from $FeO$ and $F{e}_2{O}_3$ after reaction with $Zn$ dust)
$m$-mole of $F{e}^{2+}$ from $F{e}_2{O}_3=600-125=475$
$m$ mole of $Fe{O}_3=\frac{475}{2}$
Mass of $FeO=\frac{125\times (56+16)}{1000}g=9g$
Mass of $F{e}_2{O}_3=\frac{475}{2}\times \frac{160}{1000}=38g$
$\%F{e}_2{O}_3=\frac{38}{38+9}\times 100=80.85\%$