<u>Answer:</u> The moles of precipitate (lead (II) iodide) produced is  moles
 moles
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
 .....(1)
     .....(1)
- <u>For lead (II) nitrate:</u>
Molarity of lead (II) nitrate solution = 2.70 M
Volume of solution = 33.0 mL = 0.033 L   (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:

Molarity of NaI solution = 0.00157 M
Volume of solution = 20.0 mL = 0.020 L
Putting values in equation 1, we get:

For the given chemical reaction:

By Stoichiometry of the reaction:
2 moles of NaI reacts with 1 mole of lead (II) nitrate
So,  moles of NaI will react with =
 moles of NaI will react with =  of lead (II) nitrate
 of lead (II) nitrate
As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.
Thus, NaI is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of NaI produces 1 mole of lead (II) iodide
So,  moles of NaI will produce =
 moles of NaI will produce =  of lead (II) iodide
 of lead (II) iodide
Hence, the moles of precipitate (lead (II) iodide) produced is  moles
 moles