If we attempt to react 12 grams of aluminum with 12 grams of bromine, the mass of aluminum bromide that will be produced will be

Answer 1

Let us first look at the balanced equation for this reaction:

2Al + 3Br₂ ==> 2AlBr₃ .. balanced equation

Next, we'll find the limiting reactant, as that will determine how much AlBr3 can be produced:

12 g Al x 1 mol Al / 27 g = 0.444 mols Al

12 g Br₂ x 1 mol Br₂ / 160 g = 0.075 mols Br₂

The Br₂ will be limiting since it takes 3 mols Br₂ for every 2 mols Al.

Theoretical yield of AlBr₃ = 0.075 mols Br₂ x 2 mols AlBr₃ / 3 mols Br₂ x 267 g AlBr₃/mol = 13 g AlBr₃