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

Question

asked May 22, 2023 26.7k views

1 Answer

Jack Chorley · Answer 1 · 2023-05-27T01:43:05+0000

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