Question

The balanced chemical equation for this lab is:

3CuCl2(aq) + 2Al(s) mc001-1.jpg 3Cu(s) + 2AlCl3(aq)

If 10.5 g copper chloride react with 12.4 g aluminum, what is the limiting reactant?

mc001-2.jpg

mc001-3.jpg

mc001-4.jpg

mc001-5.jpg

Answer 1

First find moles of each reactant.

Molar masses: CuCl2 = 63.55 + 2(35.45) = 134.45 g·mol−1; Al = 26.98 g·mol−1.

Moles:
- n(CuCl2) = 10.5 g / 134.45 g·mol−1 = 0.0781 mol
- n(Al) = 12.4 g / 26.98 g·mol−1 = 0.4597 mol

Stoichiometry: 3 CuCl2 react with 2 Al → 1 mol CuCl2 requires (2/3) mol Al.
Needed Al for 0.0781 mol CuCl2 = (2/3)(0.0781) = 0.0521 mol Al. You have 0.4597 mol Al, so Al is in excess and CuCl2 is limiting.

(Extra: theoretical Cu produced = 0.0781 mol × 63.55 g·mol−1 = 4.96 g. Remaining Al = 0.4597 − 0.0521 = 0.4076 mol → 0.4076 × 26.98 = 11.0 g.)

Answer: CuCl2 is the limiting reactant.