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Number of moles of $\mathrm{MnO}_4^{-}$required to oxidize one mole of ferrous oxalate completely in acidic medium will be
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Verified Answer
The correct answer is:
0.6 moles
$$
\begin{aligned}
& \left.\mathrm{FeC}_2 \mathrm{O}_4 \longrightarrow \mathrm{Fe}^{2+}+2 \mathrm{CO}_2+3 \mathrm{e}^{-}\right] \times 5 \\
& \left.\mathrm{MnO}_4{ }^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\right] \times 3 \\
& 5 \mathrm{FeC}_2 \mathrm{O}_4+3 \mathrm{MnO}_4^{-}+24 \mathrm{H}^{+} \longrightarrow 5 \mathrm{Fe}^{3+}+10 \mathrm{CO}_2+3 \mathrm{Mn}^{2+}+12 \mathrm{H}_2 \mathrm{O} \\
& 5 \mathrm{~mol} \mathrm{FeC}_2 \mathrm{O}_4 \equiv 3 \mathrm{~mol} \mathrm{MnO}_4^{-} \\
& \therefore 1 \mathrm{~mol} \mathrm{FeC}_2 \mathrm{O}_4 \equiv \frac{3}{5} \mathrm{~mol} \mathrm{MnO}_4^{-} \\
& =0.6 \mathrm{~mol} \mathrm{MnO}_4^{-} \\
&
\end{aligned}
$$
\begin{aligned}
& \left.\mathrm{FeC}_2 \mathrm{O}_4 \longrightarrow \mathrm{Fe}^{2+}+2 \mathrm{CO}_2+3 \mathrm{e}^{-}\right] \times 5 \\
& \left.\mathrm{MnO}_4{ }^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\right] \times 3 \\
& 5 \mathrm{FeC}_2 \mathrm{O}_4+3 \mathrm{MnO}_4^{-}+24 \mathrm{H}^{+} \longrightarrow 5 \mathrm{Fe}^{3+}+10 \mathrm{CO}_2+3 \mathrm{Mn}^{2+}+12 \mathrm{H}_2 \mathrm{O} \\
& 5 \mathrm{~mol} \mathrm{FeC}_2 \mathrm{O}_4 \equiv 3 \mathrm{~mol} \mathrm{MnO}_4^{-} \\
& \therefore 1 \mathrm{~mol} \mathrm{FeC}_2 \mathrm{O}_4 \equiv \frac{3}{5} \mathrm{~mol} \mathrm{MnO}_4^{-} \\
& =0.6 \mathrm{~mol} \mathrm{MnO}_4^{-} \\
&
\end{aligned}
$$
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