2 Al + 6HBr → 2AlBr3 + 3H2 Non-negotiable (must solve): What is the limiting and excess reagent if 45.6g of Al reacted with 43.7g of HBr?

Question

2 Al + 6HBr → 2AlBr3 + 3H2

Non-negotiable (must solve):

What is the limiting and excess reagent if 45.6g of Al reacted with 43.7g of HBr?

Negotiable (Solve a and one more of your choice):

a) How much mass of the excess reagent will be left after the reaction?

b) Calculate the mass of AlBr3 that will be produced after the reaction?

c) Calculate the mass of H2 produced after the reaction.

Answer 1

2 Al + 6HBr → 2AlBr₃ + 3H₂ ... balanced equation

As I mentioned in my previous answer to one of your similar questions, divided the moles or reactant by the coefficient to find the limiting reactant.

For Al: 45.6 g x 1 mol / 26.98 g = 1.69 mols Al (÷1->1.69)

For HBr: 43.7 g x 1 mol / 80.9 g = 0.540 mols HBr (÷6->0.09)

HBr is limiting

a). mols Al used = 0.54 mols HBr x 1 mol Al / 6 mol HBr = 0.09 mols Al used

mass Al used = 0.09 mols Al x 26.98 g/mol = 2.43 g Al used

mass Al left after reaction = 45.6 g - 2.43 g = 43.2 g Al left

b). mass AlBr₃ produced = 0.540 mol HBr x 2 mol AlBr₃ / 6 mol HBr x 267 g/mol = 48.1 g AlBr₃ produced

c). mass H₂ produced = 0.540 mol HBr x 3H₂ / 6 HBr x 2 g/ mol = 0.54 g H₂ produced