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(C) The balanced chemical equation for the reaction of chlorine with cold and dilute $KOH$ is:$Cl_2 + 2KOH \rightarrow KCl + KClO + H_2O$Initial moles

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$0.5 	ext{ M}, 0.5 	ext{ M}, 1 	ext{ M}$

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(C) The balanced chemical equation for the reaction of chlorine with cold and dilute $KOH$ is:$Cl_2 + 2KOH ightarrow KCl + KClO + H_2O$Initial moles of $Cl_2 = 1 	ext{ mole}$Initial moles of $KOH = 	ext{Molarity} 	imes 	ext{Volume} = 2 	ext{ M} 	imes 2 	ext{ L} = 4 	ext{ moles}$According to the stoichiometry, $1 	ext{ mole}$ of $Cl_2$ reacts with $2 	ext{ moles}$ of $KOH$.Moles of $KOH$ reacted $= 2 	ext{ moles}$Moles of $KOH$ remaining $= 4 - 2 = 2 	ext{ moles}$Moles of $Cl^{-}$ formed (from $KCl$) $= 1 	ext{ mole}$Moles of $ClO^{-}$ formed (from $KClO$) $= 1 	ext{ mole}$Final concentrations (Volume $= 2 	ext{ L}$):$[Cl^{-}] = \frac{1 	ext{ mole}}{2 	ext{ L}} = 0.5 	ext{ M}$$[ClO^{-}] = \frac{1 	ext{ mole}}{2 	ext{ L}} = 0.5 	ext{ M}$$[OH^{-}] = \frac{2 	ext{ moles}}{2 	ext{ L}} = 1 	ext{ M}$

One mole of $Cl_{2(g)}$ was passed into $2 \text{ L}$ of cold $2 \text{ M}$ $KOH$ solution. After the reaction, the concentrations of $Cl^{-}$, $ClO^{-}$ and $OH^{-}$ are respectively (assume volume remains constant):

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