Acidified potassium iodate reacts with excess potassium iodide: iodine in two different oxidation states (+5 and -1) forms iodine (0), IO3^- + I^- + H^+ -> I2 + H2O. Balance the equation and find the mass of I2 formed from 4.28 g of KIO3 with excess KI. Atomic masses: K 39, I 127, O 16. Give the answer in grams to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3rx101

Acidified potassium iodate reacts with excess potassium iodide: iodine in two different oxidation states (+5 and -1) forms iodine (0), $IO X_{3} X^{-} + I X^{-} + H X^{+} I X_{2} + H X_{2} O$ . Balance the equation and find the mass of $I X_{2}$ formed from 4.28 g of $KIO X_{3}$ with excess KI. Atomic masses: K 39, I 127, O 16. Give the answer in grams to two decimal places.

Numerical answer type, answer in g, tolerance 0.03.

The balanced equation is

IO X_{3} X^{-} + 5 I X^{-} + 6 H X^{+} 3 I X_{2} + 3 H X_{2} O

(comproportionation). Molar mass of

KIO X_{3}

= 39 + 127 + 48 = 214 g/mol, so n = 4.28/214 = 0.02 mol. Each iodate gives 3

I X_{2}

, so n(

I X_{2}

) = 0.06 mol and mass = 0.06 x 254 = 15.24 g.