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An $\pu{8 g}$ sample of $\ce{Fe3O4}$ and $\ce{Fe2O3}$ containing some inert impurity was treated with excess of aqueous $\ce{KI}$ in acidic medium, which converted all iron to $\ce{Fe^2+}$. The resulting solution was then diluted to $\pu{50 mL}$.

$\pu{10 mL}$ of it was taken and the liberated iodine required $\pu{7.2 mL}$ of $\pu{1 M}$ sodium thiosulphate to reduce all iodine.

Another $\pu{25 mL}$ was taken and the iodine was removed. The remaining solution required $\pu{4.2 mL}$ of $\pu{1 M}$ $\ce{KMnO4}$ to oxidize all $\ce{Fe^2+}$.

Calculate the percentage of composition of the mixture.

The initial solution already contains $\ce{Fe^2+}$ from $\ce{FeO}$, which does not react until the second titration.

Working backwards, milliequivalents (m.eq) of $\ce{Fe^2+}$ is $5\times 4.2\times 1 = 21$ in $\pu{25 mL}$ solution. So in $\pu{50 mL}$, the amount of $\ce{Fe^2+}$ is $42$ m.eq.

Similarly, the m.eq of iodine from titration with thiosulphate is $36$.

From the first reaction (with $\ce{KI}$), $$\text{m.eq of iodine liberated = m.eq of $\ce{Fe^2+}$ formed.}$$ The excess $\ce{Fe^2+}$ is from $\ce{FeO}$ which is present in $\ce{Fe3O4}$.

The amount of $\ce{FeO}$ is $6$ m.eq. Hence its weight is $\frac{6}{1000}\times\frac{72}{2}=\pu{0.216 g}$ since $n$-factor is $\mathrm{2}$.

Since $\pu{1 mol}$ of $\ce{Fe3O4}$ contains $\pu{72 g}$ of $\ce{FeO}$, the mass of $\ce{Fe3O4}$ is $\pu{0.6945 g}$. But this answer is wrong. Is there anything wrong in my procedure?

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