JEE Advanced 2025 Paper 2 · Chemistry · Redox Reactions and Titrations · medium

A 1.600 g mixture of FeC2O4 (molar mass 144 g mol^-1) and Fe2(C2O4)3 (molar mass 376 g mol^-1) is dissolved in dilute H2SO4 and titrated with 0.120 M…

Numerical answer

1.600\ \mathrm{g}

mixture of

\mathrm{FeC_2O_4}

(molar mass

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

) and

\mathrm{Fe_2(C_2O_4)_3}

(molar mass

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

) is dissolved in dilute

\mathrm{H_2SO_4}

and titrated with

0.120\ \mathrm{M}

\mathrm{KMnO_4}

, of which

50.0\ \mathrm{mL}

is required to reach the end point (

\mathrm{MnO_4^-\rightarrow Mn^{2+}}

). The mass percentage of

\mathrm{FeC_2O_4}

in the mixture is ____. Give the answer correct to two decimal places.

Show answer and solution

Answer: 57.27

Moles of

\mathrm{MnO_4^-}

=0.0500\times0.120=6.00\times10^{-3}

, each accepting

5

electrons, so

n_{e^-}=0.0300

. In

\mathrm{FeC_2O_4}

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

(1 e

^-

) and oxalate

\rightarrow2\mathrm{CO_2}

(2 e

^-

3

^-

per formula unit. In

\mathrm{Fe_2(C_2O_4)_3}

iron is already

+3

and only three oxalates are oxidised:

6

^-

per formula unit. Let

x

and

y

be the moles:

3x+6y=0.0300

and

144x+376y=1.600

. From the first,

x=0.0100-2y

; substituting,

1.440-288y+376y=1.600

, so

88y=0.160

y=1.818\times10^{-3}

and

x=6.364\times10^{-3}

. Mass of

\mathrm{FeC_2O_4}=6.364\times10^{-3}\times144=0.9164\ \mathrm{g}

, i.e.

0.9164/1.600=57.27\%

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