When 16.6 g of KI is treated with an excess of KCIO3 in the presence of 6M HCl, ICl is produced. The amount of KCIO3 consumed and the ICl formed are:

Explanation

First, calculate the number of moles of KI given its mass: Moles of KI = Mass (g) / Molar Mass (g/mol) Moles of KI = 16.6g / (39.1 g/mol + 126.9 g/mol) (using the molar masses of K and I) Moles of KI ≈ 0.1 mol The balanced chemical equation for the reaction is as follows: 3 KI + KCIO3 + 6 HCl → 3 ICl + 3 H2O + 4 KCl From the balanced equation, you can see that 1 mole of KIO3 is required to produce 3 moles of ICl. So, the moles of KIO3 consumed would be: KI : KClO3 3 : 1 0.1 : 0.03 Using the same equation, the moles of ICl formed are related to the moles of KI: KI : ICI 3 : 1 1 : 1 0.1 : 0.1 So, the amount of KIO3 consumed is 0.03 mol, and the amount of ICl formed is 0.1 mol. The nearest answer is (b) 0.03 mol and 0.1 mol.