JEE Advance - Chemistry (1996 - No. 2)
A 3.00 g sample containing Fe3O4, Fe2O3 and an inert impure substance is treated with excess of KI solution in presence of dilute H2SO4. The entire iron is converted into Fe2+ 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 Na2S2O3 solution to reduce the iodine present. A 50 ml of the diluted solution, after complete extraction of the iodine requires 12.80 ml of 0.25 M KMnO4 solution in the dilute H2SO4 medium for the oxidation of Fe2+. Calculate the percentages of Fe2O3 and Fe3O4 in the original sample.
34.80% Fe₃O₄ and 49.33% Fe₂O₃
49.33% Fe₃O₄ and 34.80% Fe₂O₃
25.00% Fe₃O₄ and 75.00% Fe₂O₃
75.00% Fe₃O₄ and 25.00% Fe₂O₃
66.67% Fe₃O₄ and 33.33% Fe₂O₃
Explanation
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,
$$ \begin{aligned} & \mathrm{Fe}_3 \mathrm{O}_4+\mathrm{I}^{-} \longrightarrow 3 \mathrm{Fe}^{2+}+\mathrm{I}_2\left(n \text {-factor of } \mathrm{Fe}_3 \mathrm{O}_4=2\right) \\\\ & \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2\left(n \text {-factor of } \mathrm{Fe}_2 \mathrm{O}_3=2\right) \end{aligned} $$
$\begin{aligned} \Rightarrow \text { Meq of } \mathrm{I}_2 \text { formed } & =\mathrm{Meq}\left(\mathrm{Fe}_3 \mathrm{O}_4+\mathrm{Fe}_2 \mathrm{O}_3\right) =\text { Meq of hypo required } \\\\ \Rightarrow 2 x+2 y & =11 \times 0.5 \times 5=27.5 ...........(i)\end{aligned}$
Now, total millimol of $\mathrm{Fe}^{2+}$ formed $=3 x+2 y$. In the reaction
$$ \begin{array}{ll} \mathrm{Fe}^{2+}+\mathrm{MnO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+} \\\\ n \text {-factor of } \mathrm{Fe}^{2+}=1 \\\\ \Rightarrow \text { Meq of } \mathrm{MnO}_4^{-}=\text {Meq of } \mathrm{Fe}^{2+} \\\\ \Rightarrow 3 x+2 y=12.8 \times 0.25 \times 5 \times 2=32 ...........(ii) \end{array} $$
Solving Eqs. (i) and (ii), we get
$$ \begin{aligned} x=4.5 \text { and } y=9.25 \end{aligned} $$
$\begin{aligned} \Rightarrow \text { Mass of } \mathrm{Fe}_3 \mathrm{O}_4 & =\frac{4.5}{1000} \times 232=1.044 \mathrm{~g} \\\\ \% \text { mass of } \mathrm{Fe}_3 \mathrm{O}_4 & =\frac{1.044}{3} \times 100=34.80 \% \\\\ \text { Mass of } \mathrm{Fe}_2 \mathrm{O}_3 & =\frac{9.25}{1000} \times 160=1.48 \mathrm{~g} \\\\ \% \text { mass of } \mathrm{Fe}_2 \mathrm{O}_3 & =\frac{1.48}{3} \times 100=49.33 \%\end{aligned}$
$$ \begin{aligned} & \mathrm{Fe}_3 \mathrm{O}_4+\mathrm{I}^{-} \longrightarrow 3 \mathrm{Fe}^{2+}+\mathrm{I}_2\left(n \text {-factor of } \mathrm{Fe}_3 \mathrm{O}_4=2\right) \\\\ & \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2\left(n \text {-factor of } \mathrm{Fe}_2 \mathrm{O}_3=2\right) \end{aligned} $$
$\begin{aligned} \Rightarrow \text { Meq of } \mathrm{I}_2 \text { formed } & =\mathrm{Meq}\left(\mathrm{Fe}_3 \mathrm{O}_4+\mathrm{Fe}_2 \mathrm{O}_3\right) =\text { Meq of hypo required } \\\\ \Rightarrow 2 x+2 y & =11 \times 0.5 \times 5=27.5 ...........(i)\end{aligned}$
Now, total millimol of $\mathrm{Fe}^{2+}$ formed $=3 x+2 y$. In the reaction
$$ \begin{array}{ll} \mathrm{Fe}^{2+}+\mathrm{MnO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+} \\\\ n \text {-factor of } \mathrm{Fe}^{2+}=1 \\\\ \Rightarrow \text { Meq of } \mathrm{MnO}_4^{-}=\text {Meq of } \mathrm{Fe}^{2+} \\\\ \Rightarrow 3 x+2 y=12.8 \times 0.25 \times 5 \times 2=32 ...........(ii) \end{array} $$
Solving Eqs. (i) and (ii), we get
$$ \begin{aligned} x=4.5 \text { and } y=9.25 \end{aligned} $$
$\begin{aligned} \Rightarrow \text { Mass of } \mathrm{Fe}_3 \mathrm{O}_4 & =\frac{4.5}{1000} \times 232=1.044 \mathrm{~g} \\\\ \% \text { mass of } \mathrm{Fe}_3 \mathrm{O}_4 & =\frac{1.044}{3} \times 100=34.80 \% \\\\ \text { Mass of } \mathrm{Fe}_2 \mathrm{O}_3 & =\frac{9.25}{1000} \times 160=1.48 \mathrm{~g} \\\\ \% \text { mass of } \mathrm{Fe}_2 \mathrm{O}_3 & =\frac{1.48}{3} \times 100=49.33 \%\end{aligned}$
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