Chlorine gas is passed into aqueous KI containing starch until the blue colour first appears and then disappears. The final iodine-containing product is iodate through $5 Cl X_{2} + I X_{2} + 6 H X_{2} O 2 HIO X_{3} + 10 HCl$ . The number of moles of $Cl X_{2}$ consumed per mole of $I X^{-}$ in going from $I X^{-}$ to $IO X_{3} X^{-}$ is

Answer: C

First,

Cl X_{2} + 2 I X^{-} 2 Cl X^{-} + I X_{2}

uses 0.5 mol

Cl X_{2}

per mol

I X^{-}

and gives 0.5 mol

I X_{2}

. Then 0.5 mol

I X_{2}

needs

0.5 \times 5 = 2.5

mol

Cl X_{2}

. The total is

0.5 + 2.5 = 3.0

mol. This matches electron counting:

I X^{-} \to IO X_{3} X^{-}

loses 6 electrons and each

Cl X_{2}

takes 2.