How many grams of aluminum will be produced if 12.0g of Ba react with 9.0g of Al (SO4) 3

The balanced reaction would be:

3 Ba + Al₂ (SO₄) ₃ - - > 3 BaSO₄ + 2 Al

The molar mass of Ba is 137.3 g/mol. So,

Moles Ba = 12/137.3 = 0.0874 mol Ba

The molar mass of Al₂ (SO₄) ₃ is 342.15 g/mol. So,

Moles Al (SO₄) ₃ = 9/342.15 = 0.0263 mol Al₂ (SO₄) ₃

Next, determine the limiting reactant:

0.0874 mol Ba (1 mol Al₂ (SO₄) ₃ / 3 mol Ba) = 0.0291 mol Al₂ (SO₄) ₃

Since the available moles of Al₂ (SO₄) ₃ is less than the theoretical amount needed (0.0263<0.0291), Al₂ (SO₄) ₃ is the limiting reactant.

Next, let us base the amount of Al from the limiting reactant.

0.0263 mol Al₂ (SO₄) ₃ (2 mol Al / 1 mol Al₂ (SO₄) ₃) = 0.0526 mol Al

Since the molar mass of Al is 26.98 g/mol,

Mass of Al = 0.0526*26.98 = 1.42 g Al