Question

A $3.00\, g$ sample containing $Fe_3O_4,Fe_2O_3$ and an inert impure substance, is treated with excess of $KI$ solution in presence of dilute $H_2SO_4$. The entire iron is converted into $Fe^{2+}$ along with the liberation of iodine. The resulting solution is diluted to $100 \,mL $. A $20\, mL$ of the diluted solution requires $11.0 \,mL$ of $0.5 \,M \,Na_2S_2O_3$ solution to reduce the iodine present. A $50 \,mL$ of the dilute solution, after complete extraction of the iodine required $12.80 \,mL$ of $0.25 \,M \,KMnO_4$ solution in dilute $H_2SO_4$ medium for the oxidation of $Fe^{2+}$. Calculate the percentage of $Fe_2O_3$ and $Fe_3O_4$ in the original sample.

Answer 1

Let the original sample contains $x$ millimol of $Fe _{3} O _{4}$ and $y$ millimol of $Fe _{2} O _{3}$. In the first phase of reaction,
$Fe _{3} O _{4}+ I ^{-} \longrightarrow 3 Fe ^{2+}+ I _{2}\left(n\right.$-factor of $\left.Fe _{3} O _{4}=2\right)$
$Fe _{2} O _{3}+ I ^{-} \longrightarrow 2 Fe ^{2+}+ I _{2}\left(n\right.$-factor of $\left.Fe _{2} O _{3}=2\right)$
$\Rightarrow$ Meq of $I_{2}$ formed $=$ Meq $\left( Fe _{3} O _{4}+ Fe _{2} O _{3}\right)$
= Meq of hypo required
$\Rightarrow 2 x+2 y=11 \times 0.5 \times 5=27.5....$(i)
Now, total millimol of $Fe ^{2+}$ formed $=3 x+2 y$. In the reaction
$Fe ^{2+}+ MnO _{4}^{-}+ H ^{+} \longrightarrow Fe ^{3+}+ Mn ^{2+}$
$n$-factor of $Fe ^{2+}=1$
$\Rightarrow Meq$ of $MnO _{4}^{-}= Meq$ of $Fe ^{2+}$
$\Rightarrow 3 x+2 y=12.8 \times 0.25 \times 5 \times 2=32.....$(ii)
Solving Eqs. (i) and (ii), we get
$x=4.5$
and $y=9.25$
$\Rightarrow $ Mass of $Fe _{3} O _{4}=\frac{4.5}{1000} \times 232=1.044 \,g$
$\%$ mass of $Fe _{3} O _{4}=\frac{1.044}{3} \times 100=34.80 \%$
Mass of $Fe _{2} O _{3}=\frac{9.25}{1000} \times 160=1.48\, g$
$\%$ mass of $Fe _{2} O _{3}=\frac{1.48}{3} \times 100=49.33 \%$