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(B) The balanced chemical equation for the reaction of $KMnO_4$ with $KI$ in a weakly basic or neutral medium is:$2KMnO_4 + KI + H_2O \longrightarrow

Vedclass · Accepted Answer

$6$

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(B) The balanced chemical equation for the reaction of $KMnO_4$ with $KI$ in a weakly basic or neutral medium is:$2KMnO_4 + KI + H_2O \longrightarrow 2MnO_2 + KIO_3 + 2KOH$ (Note: $I^-$ is oxidized to $IO_3^-$ in neutral/basic medium).However,if the reaction is considered as $I^-$ to $I_2$ (as implied by the question context):$2KMnO_4 + 6KI + 4H_2O \longrightarrow 2MnO_2 + 3I_2 + 8KOH$.From the stoichiometry of the balanced equation,$2$ moles of $KMnO_4$ produce $3$ moles of $I_2$.Therefore,for $4$ moles of $KMnO_4$,the moles of $I_2$ produced = $(4 	imes 3) / 2 = 6$ moles.

In the chemical reaction between stoichiometric quantities of $KMnO_4$ and $KI$ in a weakly basic or neutral solution,what is the number of moles of $I_2$ released for $4$ moles of $KMnO_4$ consumed?

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