What mass of precipitate will form if 1.50 l of highly concentrated pb(clo3)2 is mixed with 0.350 l of 0.170 m nai? assume the reaction goes to completion.

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

Because a model would make the investigation easier to understand