C. Calculate the grams of lead (II) iodide that can be reproduced from 75.00 grams of potassium iodide
75.00 g of KI = 75.00 / 166 = 0.452 moles
0.45 moles of KI produces 0.45 /2 = 0.226 mol PbI2
Amount of PbI2 produced = 0.226 x 461
= 104.1 grams
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