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

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