Some Basic Concepts of Chemistry - Result Question 34
Subjective Questions
36. A 3.00 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 mL . A 20 mL of the diluted solution requires 11.0 mL of $0.5 \mathrm{M} \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ solution to reduce the iodine present. A 50 mL of the dilute solution, after complete extraction of the iodine required 12.80 mL of $0.25 \mathrm{M} \mathrm{KMnO}_4$ 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 M)
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Solution:
Let the original sample contain $x$ millimol of $\mathrm{Fe}_3 \mathrm{O}_4$ and $y$ millimol of $\mathrm{Fe}_2 \mathrm{O}_3$. In the first phase of reaction,
$\begin{aligned} \mathrm{Fe}_3 \mathrm{O}_4+\mathrm{I}^{-} & \longrightarrow 3 \mathrm{Fe}^{2+}+\mathrm{I}_2\left(n \text {-factor of } \mathrm{Fe}_3 \mathrm{O}_4=8\right) \\ \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{I}^{-} & \longrightarrow 2 \mathrm{Fe}^{3+}+\mathrm{I}_2\left(n \text {-factor of } \mathrm{Fe}_2 \mathrm{O}_3=3\right) \\ \Rightarrow \mathrm{Meq} \text { of } \mathrm{I}_2 \text { formed } & =\mathrm{Meq}\left(\mathrm{Fe}_3 \mathrm{O}_4+\mathrm{Fe}_2 \mathrm{O}_3\right) \\ & =\text { Meq of hypo needed } \ \Rightarrow \quad 2 x+2 y & =11 \times 0.5 \times 5=27.5\hspace{15mm} …(i) \end{aligned}$
Now, total millimol of $\mathrm{Fe}^{2+}$ formed $=x+2 y$. In the reaction
$\begin{aligned} & \mathrm{Fe}^{2+}+\mathrm{MnO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+} \\ & n \text {-factor of } \mathrm{Fe}^{2+}=2 \\ & \Rightarrow \quad \text { Meq of } \mathrm{MnO}_4^{-}=\mathrm{Meq} \text { of } \mathrm{Fe}^{2+} \\ & \Rightarrow \quad 3 x+2 y=12.8 \times 0.25 \times 5 \times 2=32 \hspace{15mm} …(ii)\\ & \end{aligned}$
Solving Eqs. (i) and (ii), we get
$\begin{aligned} x & =4.5 \text { and } y=9.25 \\ \Rightarrow \quad \mathrm{Me}_3 \mathrm{O}_4 & =\frac{4.5}{1000} \times 232=1.044 \mathrm{~g} \\ % \text { mass of } \mathrm{Fe}_3 \mathrm{O}_4 & =\frac{1.044}{3} \times 100=34.80 % \\ \text { Mass of } \mathrm{Fe}_2 \mathrm{O}_3 & =\frac{9.25}{1000} \times 160=1.48 \mathrm{~g} \\ % \text { mass of } \mathrm{Fe}_2 \mathrm{O}_3 & =\frac{1.48}{1.48+1.60} \times 100=49.33 % \end{aligned}$
BETA
