Question Detail
Question
In the reaction of aluminum with iodine to form AlI₃, predict the amount of product that can be made from 0.50 mol of aluminum and 0.60 mol of iodine.
Correct Answer
Option C is the correct answer.
Detailed Explanation
The balanced equation is 2Al + 3I₂ → 2AlI₃. 0.50 mol Al requires 0.75 mol I₂ (0.50 × 3/2). Since only 0.60 mol I₂ is available, I₂ is limiting. Amount of AlI₃ = 0.60 mol I₂ × (2 mol AlI₃/3 mol I₂) = 0.40 mol.
Hint
The notes provide the answer as 0.40 mol.