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

Rezan Achmad asked Oct 5, 2023

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Rezan Achmad asked Oct 5, 2023

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13 votes

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

Vikesh Prasad answered Oct 9, 2023

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