Final answer:
To find the grams of aluminum produced from the reaction of Ba with Al₂(SO₄)₃, stoichiometry is used along with the molar masses of the reactants. The amount produced is determined by identifying the limiting reagent and using the stoichiometric ratios from the balanced equation.
Step-by-step explanation:
To determine how many grams of aluminum will be produced when 1.0 g of Ba react with 1.80 g of Al₂(SO₄)₃, we must employ stoichiometry and the given balanced chemical equation 3Ba + Al₂(SO₄)₃ → 2Al + 3BaSO₄.
Firstly, we calculate the molar masses. The molar mass of Ba is approximately 137.33 g/mol, and that of Al₂(SO₄)₃ is approximately 342.15 g/mol. Next, we find the moles of Ba and Al₂(SO₄)₃ that are reacting by dividing their respective masses by the molar masses:
Moles of Ba = 1.0 g / 137.33 g/mol
Moles of Al₂(SO₄)₃ = 1.80 g / 342.15 g/mol
Using the stoichiometry of the balanced equation, we know that 3 moles of Ba will produce 2 moles of Al. We then calculate which reactant is the limiting reagent, as it will dictate the amount of product formed. In this case, we'll assume Ba is the limiting reagent for the sake of explanation:
Moles of Al produced = (Moles of Ba) x (2 moles Al / 3 moles Ba)
However, without performing actual numeric calculations and comparisons between the moles of Ba and Al₂(SO₄)₃, we cannot conclusively determine which is the limiting reagent or calculate the exact amount of Al produced. Yet, the method described is how you would correctly approach the problem to find the mass of aluminum produced in the reaction.