How much lead (ii) iodate, pb (io3) 2 (molar mass = 557.0 g/mol), is precipitated when 3.20  102 ml of 0.285 m pb (no3) 2 (aq) are mixed with 386 ml of 0.512 m naio3 (aq) solution? pb (no3) 2 (aq) + naio3 (aq)  pb (io3) 2 (s) + nano3 (aq) ?

First, determine the moles of reactants available.

Moles Pb (NO₃) ₂: 0.285 mol/L * 1 L/1000 mL * 320 mL = 0.0912

Moles NaIO₃: 0.512 mol/L * 1 L/1000 mL * 386 mL = 0.1976

Next, determine which reactant is limiting. The balanced reaction is:

Pb (NO₃) ₂ (aq) + 2 NaIO₃ (aq) - - > Pb (IO₃) ₂ (s) + 2 NaNO₃ (aq)

Theoretical Amount of NaIO₃ needed for 0.0912 mol Pb (NO₃) ₂: 0.0912 mol Pb (NO₃) ₂ * 2 mol NaIO₃/1 mol Pb (NO₃) ₂ = 0.1824 moles NaIO₃

Since the theoretical amount is less than what's available, this means NaIO₃ is the excess reactant while Pb (NO₃) ₂ is the limiting reactant. Thus, let; s base our answer using the amount of 0.0912 mol Pb (NO₃) ₂.

Mass of Precipitate = 0.0912 mol Pb (NO₃) ₂ * 1 mol Pb (IO₃) ₂/1 mol Pb (NO₃) ₂ * 557 g/mol = 50.8 grams of Pb (IO₃) ₂ precipitate