KMnO_4 reacts with KI in basic medium to form | vedclass

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(B) Moles of $KI = 0.1 	imes 0.250 = 0.025 \ mol$. Moles of $KMnO_4 = 0.02 	imes 0.250 = 0.005 \ mol$. The balanced chemical equation is: $2MnO_4^-

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$0.0075$

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(B) Moles of $KI = 0.1 	imes 0.250 = 0.025 \ mol$. Moles of $KMnO_4 = 0.02 	imes 0.250 = 0.005 \ mol$. The balanced chemical equation is: $2MnO_4^- + 6I^- + 4H_2O ightarrow 2MnO_2 + 3I_2 + 8OH^-$. According to the stoichiometry, $2 \ moles$ of $KMnO_4$ react with $6 \ moles$ of $KI$ to produce $3 \ moles$ of $I_2$. Here, $KMnO_4$ is the limiting reagent because $0.005 \ mol$ of $KMnO_4$ requires $0.015 \ mol$ of $KI$ (which is less than the $0.025 \ mol$ available). Moles of $I_2$ formed $= \frac{3}{2} 	imes 	ext{moles of } KMnO_4 = \frac{3}{2} 	imes 0.005 = 0.0075 \ mol$.

$KMnO_4$ 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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