posted by Diggs .
How many grams of lead(II) iodate, Pb(IO3)2 (formula weight = 557.0 g/mol), are 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? The unbalanced equation of reaction is:
Pb(NO3)2(aq) + NaIO3(aq) → Pb(IO3)2(s) + NaNO3(aq)
Balance the equation.
Calculate moles of each from moles = M x L.
Determine the limiting reagent.
moles Pb(IO3)2 x molar mass = grams.
I note there is an excess of NaIO3 present which will reduce the solubility but I don't think you need to consider that.