When 16.6 g of KI is treated with an excess of KIO3 in the presence of 6NHCl, ICl is produced. The amount of KIO3 consumed and the ICl formed are:

Explanation

To find the correct amounts of KIO3 consumed and ICl formed, first, the balanced chemical equation must be established. Based on stoichiometry, the moles of KI can be determined from its mass:Molar mass of KI = 166 g/mol, thus:Moles of KI = 16.6 g / 166 g/mol = 0.1 mol.Assuming the balanced reaction shows that 1 mole of KI reacts with 1 mole of KIO3 to produce 3 moles of ICl, the limiting reactant is KI, and we would need only 0.05 mol of KIO3 (0.1 mol KI) to react completely with it. Therefore, the amount of ICl produced would be 0.15 mol (0.1 mol KI produces 0.3 mol ICl). Hence, the correct answer is Option B: 0.05 mol of KIO3 consumed and 0.15 mol of ICl formed. The other options either overestimate or underestimate the amounts based on the stoichiometric relationships established in the balanced equation.