In neutral or faintly alkaline medium,MnO_4^{ | vedclass

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Question

(D) The balanced chemical equation for the reaction in a neutral or faintly alkaline medium is:$2KMnO_4 + H_2O + KI \rightarrow 2MnO_2 + 2KOH + KIO_3$

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(D) The balanced chemical equation for the reaction in a neutral or faintly alkaline medium is:$2KMnO_4 + H_2O + KI ightarrow 2MnO_2 + 2KOH + KIO_3$From the stoichiometry of the balanced equation,$1 \ mole$ of $KI$ reacts with $2 \ moles$ of $KMnO_4$.Given,the amount of $KI = 	ext{Molarity} 	imes 	ext{Volume} = 0.5 \ M 	imes 1 \ L = 0.5 \ moles$.Since $1 \ mole$ of $KI$ requires $2 \ moles$ of $KMnO_4$,then $0.5 \ moles$ of $KI$ will require:$0.5 	imes 2 = 1.0 \ mole$ of $KMnO_4$.

In neutral or faintly alkaline medium,$MnO_4^{-}$ oxidizes $I^{-}$ to iodate. What is the number of moles of $KMnO_4$ required to completely convert $1 \ L$ of $0.5 \ M \ KI$ to iodate?

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