How many grams of Pb(IO3)2(s) are precipitated when 27.0 mL of 3.00 M Pb(NO3)2(aq) are mixed with 30.0 mL of a 5.00 M KIO3(aq) solution?

Question

3.00M(.o27)(1mol Pb(IO3)2)/1mol Pb(NO3)2)(557/1)

Answer 1

This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
Pb(NO3)2 + 2KIO3 ==> Pb(IO3)2 + 2KNO3
mols Pb(NO3)2 = M x L = ?
mols KIO3 = M x L = ?

Using the coefficients in the balanced equation, convert mols Pb(NO3)2 to mols Pb(IO3)2 and I think you have done that.
Do the same for mols KIO3 to mols Pb(IO3)2. It is likely the two answers will not be the same which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Using the smaller value convert mols Pb(IO3)2 to g by grams = mols x molar mass. That is the theoretical yield for Pb(IO3)2 and it assumes everything goes at 100% (no losses).

Answer 2

45.1