1 mole of MnO_4^{2-} in neutral aqueous medi | vedclass

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Question

(A) The disproportionation reaction of manganate ion $(MnO_4^{2-})$ in a neutral aqueous medium is given by: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^-

Vedclass · Accepted Answer

$2/3 \ mole$ of $MnO_4^-$ and $1/3 \ mole$ of $MnO_2$

Vedclass · Answer

(A) The disproportionation reaction of manganate ion $(MnO_4^{2-})$ in a neutral aqueous medium is given by: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$ From the stoichiometry of the balanced equation,$3 \ moles$ of $MnO_4^{2-}$ produce $2 \ moles$ of $MnO_4^-$ and $1 \ mole$ of $MnO_2$. Therefore,$1 \ mole$ of $MnO_4^{2-}$ will produce $\frac{2}{3} \ mole$ of $MnO_4^-$ and $\frac{1}{3} \ mole$ of $MnO_2$.

$1 \ mole$ of $MnO_4^{2-}$ in neutral aqueous medium disproportionates to:

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