Answer : The mass of  precipitate produced will be, 9.681 grams.
 precipitate produced will be, 9.681 grams.
Explanation : Given,
Molarity of NaI = 0.210 M
Volume of solution = 0.2 L
Molar mass of  = 461.01 g/mole
 = 461.01 g/mole
First we have to calculate the moles of  .
.
 
Now we have to calculate the moles of  .
.
The balanced chemical reaction is,
 
From the balanced reaction we conclude that
As, 2 moles of  react to give 1 mole of
 react to give 1 mole of  
So, 0.042 moles of  react to give
 react to give  moles of
 moles of  
Now we have to calculate the mass of  .
.
 
 
Therefore, the mass of  precipitate produced will be, 9.681 grams.
 precipitate produced will be, 9.681 grams.