Chemistry stoichiometry

Question

15.3 L of Fluorine gas reacts with 35.5 g of Aluminium chloride. how many Liters of gaseous product will be produced under STP conditions .

Answer 1

3F₂(g) + 2AlCl₃(s) ==> 2AlF₃(s) + 3Cl₂(g)

Under standard condition (STP), 1 mole of gas = 22.4 L.

moles F₂ gas used = 15.3 L x 1 mol / 22.4 L = 0.683 mols F₂ gas

moles AlCl₃ used = 35.5 g x 1 mol AlCl₃ / 133.3 g = 0.266 mols AlCl₃

Since the mol ratio of AlCl₃ : F₂ is 2 : 3, AlCl₃ is the limiting reactant as it will run out first.

Moles Cl₂ formed = 0.266 mols AlCl₃ x 3 mols Cl₂ / 2 mols AlCl₃ = 0.399 mols Cl₂(g)

Liters of Cl₂(g) = 0.399 mols Cl₂ x 22.4 L / 1 mol = 8.94 L Cl₂