<u>Answer:</u> The mass of lead iodide produced is 9.22 grams
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

Molarity of NaI = 0.200 M
Volume of solution = 0.200 L
Putting values in above equation, we get:

The chemical equation for the reaction of NaI and lead chlorate follows:

By Stoichiometry of the reaction:
2 moles of NaI reacts produces 1 mole of lead iodide
So, 0.04 moles of NaI will react with =
of lead iodide
To calculate the number of moles, we use the equation:
Molar mass of lead iodide = 461 g/mol
Moles of lead iodide= 0.02 moles
Putting values in above equation, we get:

Hence, the mass of lead iodide produced is 9.22 grams