Some Basic Concepts of Chemistry 2 Question 36
36. A $3.00 \mathrm{~g}$ sample containing $\mathrm{Fe}{3} \mathrm{O}{4}, \mathrm{Fe}{2} \mathrm{O}{3}$ and an inert impure substance, is treated with excess of KI solution in presence of dilute $\mathrm{H}{2} \mathrm{SO}{4}$. The entire iron is converted into $\mathrm{Fe}^{2+}$ along with the liberation of iodine. The resulting solution is diluted to $100 \mathrm{~mL}$. A $20 \mathrm{~mL}$ of the diluted solution requires $11.0 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{Na}{2} \mathrm{~S}{2} \mathrm{O}{3}$ solution to reduce the iodine present. A $50 \mathrm{~mL}$ of the dilute solution, after complete extraction of the iodine required $12.80 \mathrm{~mL}$ of $0.25 \mathrm{M} \mathrm{KMnO}{4}$ solution in dilute $\mathrm{H}{2} \mathrm{SO}{4}$ medium for the oxidation of $\mathrm{Fe}^{2+}$. Calculate the percentage of $\mathrm{Fe}{2} \mathrm{O}{3}$ and $\mathrm{Fe}{3} \mathrm{O}{4}$ in the original sample.
$(1996,5 \mathrm{M})$
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Solution:
- Let the original sample contains $x$ millimol of $\mathrm{Fe}{3} \mathrm{O}{4}$ and $y$ millimol of $\mathrm{Fe}{2} \mathrm{O}{3}$. In the first phase of reaction,
$$ \begin{gathered} \mathrm{Fe}{3} \mathrm{O}{4}+\mathrm{I}^{-} \longrightarrow 3 \mathrm{Fe}^{2+}+\mathrm{I}{2}\left(n \text {-factor of } \mathrm{Fe}{3} \mathrm{O}{4}=2\right) \ \mathrm{Fe}{2} \mathrm{O}{3}+\mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}{2}\left(n \text {-factor of } \mathrm{Fe}{2} \mathrm{O}{3}=2\right) \end{gathered} $$
$\Rightarrow$ Meq of $\mathrm{I}{2}$ formed $=\mathrm{Meq}\left(\mathrm{Fe}{3} \mathrm{O}{4}+\mathrm{Fe}{2} \mathrm{O}_{3}\right)$
$=$ Meq of hypo required
$$ \Rightarrow \quad 2 x+2 y=11 \times 0.5 \times 5=27.5 $$
Now, total millimol of $\mathrm{Fe}^{2+}$ formed $=3 x+2 y$. In the reaction
$$ \mathrm{Fe}^{2+}+\mathrm{MnO}_{4}^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+} $$
$n$-factor of $\mathrm{Fe}^{2+}=1$
$$ \begin{array}{ll} \Rightarrow & \text { Meq of } \mathrm{MnO}_{4}^{-}=\text {Meq of } \mathrm{Fe}^{2+} \ \Rightarrow & 3 x+2 y=12.8 \times 0.25 \times 5 \times 2=32 \end{array} $$
Solving Eqs. (i) and (ii), we get
$$ \begin{aligned} x & =4.5 \text { and } y=9.25 \ \Rightarrow \quad \text { Mass of } \mathrm{Fe}{3} \mathrm{O}{4} & =\frac{4.5}{1000} \times 232=1.044 \mathrm{~g} \ % \text { mass of } \mathrm{Fe}{3} \mathrm{O}{4} & =\frac{1.044}{3} \times 100=34.80 % \ \text { Mass of } \mathrm{Fe}{2} \mathrm{O}{3} & =\frac{9.25}{1000} \times 160=1.48 \mathrm{~g} \ \text { % mass of } \mathrm{Fe}{2} \mathrm{O}{3} & =\frac{1.48}{3} \times 100=49.33 % \end{aligned} $$
