13.7 g
The balanced formula is:
Pb(ClO3)2(aq)+2NaI(aq) ==> PbI2(s) + 2NaClO3(aq)
The number of moles of NaI we have is the volume of fluid multiplied by the molarity so
0.350 * 0.170 = 0.0595 moles
Since the NaI is the limiting reactant, for every two moles used, we'll produce 1 mole of precipitate. So
0.0595 mole / 2 = 0.02975 mole
Now we need to calculate the molar mass of PbI2. Looking up the atomic weights
Atomic weight Lead = 207.2
Atomic weight iodine = 126.90447
Molar mass PbI2 = 207.2 + 2 * 126.90447 = 461.00894 g/mol
Now multiply the molar mass by the number of moles we have.
461.00894 g/mol * 0.02975 mol = 13.71501597 g
Rounded to 3 significant figures, the answer is 13.7 g
