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