An easy way to find the limiting reactant is to divide the moles of each reactant by its coefficient in the balanced equation and whichever value is less is the limiting reactant.
Reaction: 2NaCl(aq) + Pb(NO3)2(aq)=> 2NaNO3(aq) + PbCl2(s) ... balanced equation
For NaCl: 60 g NaCl x 1 mol NaCl / 58.4 g = 1.03 mols (÷2 -> 0.51)
For Pb(NO3)2: 55 g Pb(NO3)2 x 1 mol / 331 g = 0.167 mols (÷1 -> 0.167)
Limiting reactant = Pb(NO3)2 @ 0.167 mols
Excess reactant = NaCl @ 1.03 mols
To find amount of NaCl left over, find amount used up, and then subtract that from amount initially present.
Amount NaCl used up: 0.167 mols Pb(NO3)2 x 2 mols NaCl / mol Pb(NO3)2 = 0.332 mols used up
Amount left over = 1.03 mols - 0.332 mols = 0.698 mols NaCl left over
Grams left over = 0.698 mols NaCl x 58.4 g/mol = 40.8 grams (ANSWER A)