JEE Main 2024 · Chemistry · Redox Reactions and Titrations · medium

120 mL of 0.0200 M KMnO4 in acidic medium just oxidises a sample of FeC2O4 (taking its molar mass as 144 g mol^-1) completely to Fe^3+ and CO2. The…

Numerical answer

120\ \mathrm{mL}

0.0200\ \mathrm{M}

\mathrm{KMnO_4}

in acidic medium just oxidises a sample of

\mathrm{FeC_2O_4}

(taking its molar mass as

144\ \mathrm{g\,mol^{-1}}

) completely to

\mathrm{Fe^{3+}}

and

\mathrm{CO_2}

. The mass of

\mathrm{FeC_2O_4}

in the sample, in mg, is

Show answer and solution

Answer: 576

\mathrm{MnO_4^-}

takes

5e^-

(

\mathrm{Mn^{VII}\to Mn^{2+}}

0.0200\times120=2.40

mmol, so

12.0

mmol of electrons. Each

\mathrm{FeC_2O_4}

gives up

1e^-

(

\mathrm{Fe^{2+}\to Fe^{3+}}

) plus

2e^-

(

\mathrm{C_2O_4^{2-}\to2CO_2}

), i.e.

3e^-

. Moles

=12.0/3=4.00

mmol, and mass

=4.00\times144=576\ \mathrm{mg}

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