The balanced equation for the reaction of aqueous Pb (ClO3)2 with aqueous NaI is:
Pb (ClO3)2 (aq) + 2NaI (aq) ⟶ PbI2 (s) + 2NaClO3 (aq)
To calculate the mass of precipitate that will form if 1.50 L of concentrated Pb (ClO3)2 is mixed with 0.800 L of 0.170 M NaI, we need to first calculate the number of moles of each reactant.
Number of moles of Pb(ClO3)2 = Concentration x Volume = 1.50 L x 6.00 M = 9.00 mol
Number of moles of NaI = Concentration x Volume = 0.800 L x 0.170 M = 0.136 mol
From the balanced equation, we can see that one mole of Pb(ClO3)2 reacts with two moles of NaI to form one mole of PbI2.
Therefore, the limiting reactant is NaI since it has fewer moles than Pb(ClO3)2.
The number of moles of PbI2 formed will be equal to half the number of moles of NaI used since one mole of NaI reacts with half a mole of Pb(ClO3)2 to form one mole of PbI2.
Number of moles of PbI2 formed = 0.5 x Number of moles of NaI used = 0.5 x 0.136 mol = 0.068 mol
The molar mass of PbI2 is approximately 461 g/mol.
Mass of precipitate formed = Number of moles x Molar mass = 0.068 mol x 461 g/mol ≈ 31.5 g.
Therefore, approximately 31.5 g of precipitate will form if the reaction goes to completion.