I think you are on the right track. You just have another step or two to go. And you're right, this is a limiting reagent problem; I usually work these the long way.
mols Pb(NO3)2 = M x L = approx 0.00121
mols Na2SO4 = approx 0.00130
Pb(NO3)2 x Na2SO4 ==> PbSO4 + 2NaNO3
Next, I determine the limiting reagent.
I convert 0.00121 mol Pb(NO3)2 to PbSO4 ASSUMING I have all of the Na2SO4 needed. Use the coefficients in the balanced equation to do that. That's
0.00121 x [1 mol PbSO4/1 mol Pb(NO3)] = 0.00121 mols PbSO4.
Then I convert 0.00130 mols Na2SO4 to mols PbSO4 ASSUMING I have all of the Pb(NO3)2 needed. That's
0.00130 mol Na2SO4 x (1 mol PbSO4/1 mol Na2SO4) = 0.00130 mols PbSO4.
Obviously, both answers can't be right. The correct value in limiting reagent problems is ALWAYS the smaller number and the reagent responsible for that is the limiting reagent.
Therefore, 0.00121 mols PbSO4 will be formed. Convert to grams by g = mols x molar mass = ?