Stoichiometry: Calculations with Chemical Formulas and Equations

70) 2Al(OH)₃(s) + 3H₂SO₄(aq) Al₂(SO₄)₃(aq) + 6H₂O(l)

       n = 0.450 mol Al(OH)₃          n = 0.550 mol H₂SO₄

       n = 0.450 mol Al(OH)₃ * 1 mol Al₂(SO₄)₃/2 mol Al(OH)₃ = 0.225 mol Al₂(SO₄)₃

       n = 0.550 mol H₂SO₄ * 1 mol Al₂(SO₄)₃/3 mol H₂SO₄ = 0.183 mol Al₂(SO₄)₃  

       H₂SO₄ produces the least amount of Al₂(SO₄)₃, therefore it is the limiting reactant.

       n = 0.183 mol Al₂(SO₄)₃ * 2 mol Al(OH)₃/1 mol Al₂(SO₄)₃ = 0.367 mol Al(OH)₃

       n = 0.450 mol Al(OH)₃ - 0.367 mol Al(OH)₃ = 0.083 mol Al(OH)₃ remains.