Answer:
From the balanced equation, we can see that the mole ratio of KI to PbI2 is 2:1. This means that for every 2 moles of KI, 1 mole of PbI2 is produced. Therefore, when 0.455 moles of potassium iodide react, approximately 104.7 grams of lead (II) iodide will be produced.
