0.10 mol of KI in water is treated with just enough chlorine to convert all the iodide into KIO3 (chlorine is reduced to Cl-). The number of moles of Cl2 required is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ph102

$0.10$ mol of $KI$ in water is treated with just enough chlorine to convert all the iodide into $KIO X_{3}$ (chlorine is reduced to $Cl X^{-}$ ). The number of moles of $Cl X_{2}$ required is

Iodine goes from -1 in

I X^{-}

to +5 in

IO X_{3} X^{-}

, a loss of 6 electrons per iodide, i.e.

0.60

mol electrons for

0.10

mol

KI

. Each

Cl X_{2}

accepts 2 electrons, so

0.60/2 = 0.30

mol

Cl X_{2}

. (

KI + 3 Cl X_{2} + 3 H X_{2} O KIO X_{3} + 6 HCl