KMnO_4 reacts with KI in basic medium to form | vedclass

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(B) The balanced redox reaction in basic medium is: $2MnO_4^{-} + 6I^{-} + 4H_2O \rightarrow 2MnO_2 + 3I_2 + 8OH^{-}$ Calculate the initial moles of r

Vedclass · Accepted Answer

$0.0075$

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(B) The balanced redox reaction in basic medium is: $2MnO_4^{-} + 6I^{-} + 4H_2O ightarrow 2MnO_2 + 3I_2 + 8OH^{-}$ Calculate the initial moles of reactants: Moles of $MnO_4^{-} = 0.02 \ M 	imes 0.250 \ L = 0.005 \ mol$ Moles of $I^{-} = 0.1 \ M 	imes 0.250 \ L = 0.025 \ mol$ According to the stoichiometry,$2 \ mol$ of $MnO_4^{-}$ reacts with $6 \ mol$ of $I^{-}$. For $0.005 \ mol$ of $MnO_4^{-}$,the required $I^{-}$ is $0.005 	imes (6/2) = 0.015 \ mol$. Since we have $0.025 \ mol$ of $I^{-}$,$MnO_4^{-}$ is the limiting reagent. From the stoichiometry,$2 \ mol$ of $MnO_4^{-}$ produces $3 \ mol$ of $I_2$. Therefore,$0.005 \ mol$ of $MnO_4^{-}$ will produce: Moles of $I_2 = 0.005 	imes (3/2) = 0.0075 \ mol$.

$KMnO_4$ reacts with $KI$ in basic medium to form $I_2$ and $MnO_2$. When $250 \ mL$ of $0.1 \ M \ KI$ solution is mixed with $250 \ mL$ of $0.02 \ M \ KMnO_4$ in basic medium,what is the number of moles of $I_2$ formed?

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