Question

When aqueous solutions of lead(II) ion are treated with potassium chromate solution, a bright yellow precipitate of lead(II) chromate, PbCrO4, forms. How many grams of lead chromate form when a 1.60-g sample of Pb(NO3)2 is added to 24.0 mL of 1.01 M K2CrO4 solution?

Answer 1

I don't see a Ksp given for PbCrO4; therefore, I will assume this is just another limiting reagent problem. I work these the long way.
mols Pb(NO3)2 = grams/molar mass
mols K3CrO4 = M x L = ?

Using the coefficients in the balanced equation, convert mols Pb(NO3)2 to mols PbCrO4.
Do the same and convert mols K2CrO4 to mols PbCrO4.
It is likely that the two values will not be the same; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent responsible for that is the limiting reagent.
Using the smaller number of mols of PbCrO4, convert to grams. g = mols x molar mass.