Limiting Reactants

Question

1. If 132.0 grams of copper(II) chloride reacts with 160.0 grams of sodium nitrate, how many grams of sodium chloride can be formed?

c. What is the limiting reactant of this reaction?

d. How many grams of copper(II) nitrate are formed?

e. How many grams of the excess reactant is left over in this reaction?

f. If 78.00 grams of sodium chloride are formed after the reaction, what is the percent yield of this reaction?

2. If 11.00 grams of lead(II) nitrate react with 13.00 grams of sodium iodide, how many grams of sodium nitrate can be formed?

c. What is the limiting reactant of this reaction?

d. How many grams of lead(II) iodide are formed?

e. How many grams of the excess reactant is left over in this reaction?

f. If 5.000 grams of sodium nitrate are formed after the reaction, what is the percent yield of this reaction?

Answer 1

There are a couple of ways of finding the limiting reactant. The one I like best is to take the moles of each reactant, and divide them by the corresponding coefficient in the balanced equation. Whichever value is less is the limiting reactant. See below for examples.

CuCl₂ + 2NaNO₃ ==> 2NaCl + Cu(NO₃)₂ ... balanced equation

30.0 g.....160.0 g..........? g..........

Limiting reactant:

CuCl2: 30.0 g x 1 mol / 134 g = 0.224 mols CuCl₂ (÷1->0.224)

NaNO3: 160 g x 1 mol / 85.0 g = 1.88 mols NaNO3 (÷2->0.94)

(c) CuCl2 is the limiting reactant since 0.224 is less than 0.94. CuCl2 will determine the amount of product that can be formed.

(d) 0.224 mol CuCl2 x 1 mol Cu(NO3)2 / mol CuCl2 x 188 g / mol Cu(NO3)2 = 42.1 g Cu(NO3)2

(e) 0.224 mol CuCl2 x 2 mol NaNO3 / mol CuCl2 = 0.448 mols NaNO3 used

1.88 mol - 0.448 = 1.43 mol NaNO3 left over

1.43 mol NaNO3 x 85.0 g/mol = 122 g NaNO3 left over

(f) percent yield = actual yield / theoretical yield

theoretical yield of NaCl = 0.224 mol CuCl2 x 2 mol NaCl / mol CuCl2 x 58.4 g NaCl/mol = 26.2 g NaCl

This being the case, it's impossible to obtain 78.00 g NaCl. This would be >100% yield.

Pb(NO₃)₂ + 2NaI ==> PbI₂ + 2NaNO₃

Follow the same procedure as above.