2Al + 3F -> 2AlF3 What will be present after the reaction and how much when 40g of aluminium and 57g of fluorine react?

Answer 1

2 Al + 3F ==> 2AlF₃ is not a balanced equation. You don't have the same number of atoms on each side. In addition, this reaction isn't feasible because F doesn't exist as such.

Perhaps, the reaction is supposed to be

2Al + 3F₂ ==> 2AlF₃

First, we must find the limiting reactant, if any.

moles Al present = 40 g Al x 1 mol Al / 27 g = 1.48 mol Al

moles F₂ present = 57 g F₂ x 1 mol F2 / 38 g = 1.50 mol F₂

Since we need 3 mol F₂ for every 2 mol Al, we don't have enough so F₂ is limiting.

Amount AlF₃ formed:

1.50 mol F₂ x 2 mol AlF₃ / 3 mol F₂ x 84 g AlF₃ / mol = 84 g AlF₃ formed

Amount Al present after the reaction:

1.50 mol F₂ x 2 mol Al / 3 mol F₂ = 1 mol Al used up

1.48 mol Al - 1.0 = 0.48 mol Al left over

0.48 mol Al x 27 g/mol = 13 g Al left over after the reaction

No F₂ left over after the reaction