1 text{ mole} of MnO_4^{2-} disproportionates | vedclass

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(A) The disproportionation reaction of manganate ion $(MnO_4^{2-})$ is given by:$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$From the stoi

Vedclass · Accepted Answer

$66.67$

Vedclass · Answer

(A) The disproportionation reaction of manganate ion $(MnO_4^{2-})$ is given by:$3MnO_4^{2-} + 4H^+ ightarrow 2MnO_4^- + MnO_2 + 2H_2O$From the stoichiometry of the balanced equation,$3 	ext{ moles}$ of $MnO_4^{2-}$ produce $2 	ext{ moles}$ of $MnO_4^-$.Therefore,$1 	ext{ mole}$ of $MnO_4^{2-}$ produces $2/3 	ext{ moles}$ of $MnO_4^-$.The mass of $Mn$ in $2/3 	ext{ moles}$ of $MnO_4^-$ is $(2/3) 	imes 55 \ g$.The total mass of $Mn$ in $1 	ext{ mole}$ of $MnO_4^{2-}$ is $1 	imes 55 \ g$.The percentage by mass of $Mn$ converted to $MnO_4^-$ is:$\frac{(2/3) 	imes 55}{55} 	imes 100 = \frac{2}{3} 	imes 100 = 66.67\%$.

$1 \text{ mole}$ of $MnO_4^{2-}$ disproportionates into $MnO_4^-$ and $MnO_2$ in an acidic medium. Calculate the percentage by mass of $Mn$ converted to $MnO_4^-$ (Molar mass of $Mn = 55 \ g/mol$).

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