When lead(II) nitrate reacts with sodium iodide, what is the correct formula for the product lead(II) iodide?Pb 2Pb22Pol4Pb214Pb2Hoo.

Lead (II) nitrate = Pb(NO3)2
Sodium iodide = NaI
When they react:
Pb(NO3)2 + 2NaI => 2NaNO3 + PbI2
PbI2 is formed as a product. It is called lead (II) iodide
Answer: PbI2