Answer to Question #112487 in General Chemistry for ashley

Question

When aqueous lead (II) nitratre reacts with aqueous potassium iodide, a bright yellow solid is formed, lead (II) iodide. If 25.0 mL of 0.728M of potassum iodide is reacted with excess lead (II) nitrate, how many grams of lead (II) iodide would be formed?

Answer 1

25.0 mL / 1000 * 0.728M = 0.0182 mol of potassium iodide.

Each 2 mol (KI) produce 1 mol (PbI)

So n (PbI) = 0.0182 mol / 2 = 0,0091 mol

M (PbI) = 334.10 g/mol

m (PbI) = 0,0091 * 334.10 = 3,04 g