If you have 26 g of aluminum and 26 g of chlorine, what is the limiting reactant?

Question

Aluminum + chlorine—> aluminum chloride

Answer 1

The balanced chemical reaction is

2 Al + 3 Cl₂ --> 2 AlCl₃

So you have 26 g of each reactant (Al and Cl₂). You cannot tell from this what the limiting reactant would be. The limiting reactant is the reactant that will make the LEAST amount of product.

Therefore, you'll need to convert each amount of reactant into moles, and then calculate how much product will be made from that amount.

26 g Al × (1 mol Al / 26.98 g Al) = 0.9638 mol Al

0.9638 mol Al × (2 mol AlCl₃ / 2 mol Al) = 0.9637 mol AlCl₃

26 g Cl₂ × (1 mol Cl₂ / 70.91 g Cl₂) = 0.3667 mol Cl₂

0.3667 mol Cl₂ × (2 mol AlCl₃ / 3 mol Cl₂) = 0.2444 mol AlCl₃

Since the Cl₂ produces the LEAST amount of product (AlCl₃), that means that Cl₂ is the limiting reactant in this problem. Cl₂ will run out as soon as the 26 g are used up, and there will be more Al than is needed.

You can easily find the theoretical yield of AlCl₃ from here by converting 0.2444 mol AlCl₃ into grams.

Hope this helps!

:) Jen