Question

A $3.00g$ sample containing $F{e}_3{O}_4,F{e}_2{O}_3$ and an inert impure substance, is treated with excess of $KI$ solution in presence of dilute $H_{2}S{O}_4$. The entire iron is converted into $F{e}^{2+}$ along with the liberation of iodine. The resulting solution is diluted to $100mL$. A $20mL$ of the diluted solution requires $11.0mL$ of $0.5MN{a}_2{S}_2{O}_3$ solution to reduce the iodine present. A $50mL$ of the dilute solution, after complete extraction of the iodine required $12.80mL$ of $0.25MKMn{O}_4$ solution in dilute $H_{2}S{O}_4$ medium for the oxidation of $F{e}^{2+}$. Calculate the percentage of $F{e}_2{O}_3$ and $F{e}_3{O}_4$ in the original sample.

Answer 1

Let the original sample contains $x$ millimol of $F{e}_3{O}_4$ and $y$ millimol of $F{e}_2{O}_3$. In the first phase of reaction,
$F{e}_3{O}_4+I^{-}⟶3F{e}^{2+}+I_2(n$-factor of $F{e}_3{O}_4=2)$
$F{e}_2{O}_3+I^{-}⟶2F{e}^{2+}+I_2(n$-factor of $F{e}_2{O}_3=2)$
$\Rightarrow$ Meq of $I_2$ formed $=$ Meq $\left(F{e}_3{O}_4+F{e}_2{O}_3\right)$
= Meq of hypo required
$\Rightarrow 2x+2y=11\times 0.5\times 5=\mathrm{27.5....}$(i)
Now, total millimol of $F{e}^{2+}$ formed $=3x+2y$. In the reaction
$F{e}^{2+}+Mn{O}_4^{-}+H^{+}⟶F{e}^{3+}+M{n}^{2+}$
$n$-factor of $F{e}^{2+}=1$
$\Rightarrow Meq$ of $Mn{O}_4^{-}=Meq$ of $F{e}^{2+}$
$\Rightarrow 3x+2y=12.8\times 0.25\times 5\times 2=\mathrm{32.....}$(ii)
Solving Eqs. (i) and (ii), we get
$x=4.5$
and $y=9.25$
$\Rightarrow$ Mass of $F{e}_3{O}_4=\frac{4.5}{1000}\times 232=1.044g$
$\%$ mass of $F{e}_3{O}_4=\frac{1.044}{3}\times 100=34.80\%$
Mass of $F{e}_2{O}_3=\frac{9.25}{1000}\times 160=1.48g$
$\%$ mass of $F{e}_2{O}_3=\frac{1.48}{3}\times 100=49.33\%$