science problem

Question

How many grams of excess reactant is left over when 60 grams of NaCl reacts with 55 grams of Pb(NO3)2?

A) 40 g

B) 60 g

C) 50 g

D) 30 g

Answer 1

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)