If 16.5 g of Al are reacted with 39.2 g of chlorine gas, then aluminum chloride is formed.

Question

1. Which reactant is the limiting reactant?
2. What is the theoretical yield?

 

Formula: 2 Al + 3 Cl2 = 2 AlCl3 (already balanced)

 

Show all work for the two questions please

Answer 1

2 Al + 3 Cl₂ ==> 2 AlCl₃  Balanced Equation

moles Al present = 16.5 g Al x 1 mole Al/26.98 g = 0.612 moles Al

moles Cl₂ prsent = 39.2 g Cl₂ x 1 mole Cl₂/35.45 g = 1.11 moles Cl₂

Dividing each reactant by it's coefficient in the balanced equation one obtains:

0.612 moles Al/2 = 0.306 mols Al

1.11 moles Cl₂/3 = 0.37 moles Cl₂

LIMITING REACTANT IS Al

 

Theoretical Yield will thus depend on moles of Al and the mole ratio of AlCl3 : Al which is 1 : 1 (see balanced equation)

0.306 moles Al x 2 moles AlCl₃/2 moles Al = 0.306 moles AlCl₃ = theoretical yield (in moles)

0.306 moles AlCl₃ x 133.3 g/mole = 40.8 g AlCl₃ = theoretical yield (in grams)