JEE Main 2016 · Chemistry · d- and f-Block Elements and Metallurgy · hard

31.6 g of KMnO4 (M=158) is heated strongly until it decomposes completely into K2MnO4, MnO2 and O2. The solid manganate is then dissolved in water…

Single correct option

31.6\ \mathrm{g}

\mathrm{KMnO_4}

(

M=158

) is heated strongly until it decomposes completely into

\mathrm{K_2MnO_4}

\mathrm{MnO_2}

and

\mathrm{O_2}

. The solid manganate is then dissolved in water and acidified so that it disproportionates completely to

\mathrm{MnO_4^-}

and

\mathrm{MnO_2}

. The mass of

\mathrm{KMnO_4}

regenerated is nearly

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Answer: (D) $10.5\ \mathrm{g}$

31.6/158=0.200

mol. Heating:

\mathrm{2KMnO_4\to K_2MnO_4+MnO_2+O_2}

, giving

0.100

mol

\mathrm{K_2MnO_4}

. In acid, manganate(VI) disproportionates:

\mathrm{3MnO_4^{2-}+4H^+\to2MnO_4^-+MnO_2+2H_2O}

, so

3

mol manganate give

2

mol permanganate. Permanganate formed

=0.100\times2/3=0.0667

mol, mass

=0.0667\times158=10.5

g. Taking all manganate to permanganate would give

15.8

g, and

21.1

g results from applying the factor

2/3

to the original

31.6

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