JEE Main 2017 · Chemistry · d- and f-Block Elements and Metallurgy · medium

44 g of FeC2O4 (M=144) is oxidised by KMnO4, once in acidic medium (to Mn^2+) and once in neutral medium (to MnO2), the Fe^2+ going to Fe^3+ and…

Single correct option

1.44\ \mathrm{g}

\mathrm{FeC_2O_4}

(

M=144

) is oxidised by

\mathrm{KMnO_4}

, once in acidic medium (to

\mathrm{Mn^{2+}}

) and once in neutral medium (to

\mathrm{MnO_2}

), the

\mathrm{Fe^{2+}}

going to

\mathrm{Fe^{3+}}

and oxalate to

\mathrm{CO_2}

. The ratio of the volumes of the same

\mathrm{KMnO_4}

solution consumed (acidic : neutral) is

Show answer and solution

Answer: (A) $3:5$

Per mole of

\mathrm{FeC_2O_4}

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

gives

1\ e^-

and

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

gives

2\ e^-

, a total of

3\ e^-

. Acidic:

\mathrm{MnO_4^-}

takes

5\ e^-

(n-factor 5), so moles

\mathrm{KMnO_4}=0.01\times3/5=0.006

. Neutral: n-factor

3

, so moles

=0.01\times3/3=0.010

. Ratio

=0.006:0.010=3:5

(also equal to the inverse ratio of n-factors,

3/5

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