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In neutral or faintly alkaline medium, $\mathrm{MnO}_4^{-}$oxidizes $\mathrm{I}^{-}$to iodate. What is the number of moles of $\mathrm{KMnO}_4$ required to completely convert $1 \mathrm{~L}$ of $0.5 \mathrm{M} \mathrm{KI}$ to iodate?
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The correct answer is:
$1.0$
Thus, according to the balanced equation, 1 mole of $\mathrm{KI}$ or $\mathrm{I}^{-}$requires 2 moles of $\mathrm{KMnO}_4$ or $\mathrm{MnO}_4^{-}$.
Therefore, $1 \times 0.5=0.5$ moles of $\mathrm{I}^{-}$would require $2 \times 0.5=1.0$ moles of $\mathrm{KMnO}_4$.
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