When lead nitrate reacts with potassium iodide, yellow precipitate ofPbI2 is formedKNO3 is formedPb(NO3)2 is formedPbIO3 is formed

When lead nitrate (Pb(NO3)2) reacts with potassium iodide (KI), a double displacement reaction occurs. Here's the step by step process:

1. Identify the reactants which are lead nitrate and potassium iodide.

2. Write down the chemical reaction: Pb(NO3)2 + 2KI → PbI2 + 2KNO3

3. In this reaction, the lead (Pb) from lead nitrate combines with the iodide (I) from potassium iodide to form lead iodide (PbI2), which is a yellow precipitate.

4. Simultaneously, the potassium (K) from potassium iodide combines with the nitrate (NO3) from lead nitrate to form potassium nitrate (KNO3).

5. So, the products of this reaction are lead iodide (PbI2) and potassium nitrate (KNO3).

Therefore, the statement "When lead nitrate reacts with potassium iodide, yellow precipitate of PbI2 is formed and KNO3 is formed" is correct. The other statements are incorrect as Pb(NO3)2 is a reactant (not a product) and PbIO3 is not formed in this reaction.