0.450 mol of aluminum hydroxide is allowed to react with 0.550 mol of sulfuric acid; the reaction which ensues is: 2al(oh)3(s) + 3h2so4(aq) > al2(so4)3(aq) + 6h2o(l) how many moles of h2o can form under these conditions?

The answer to your question is 1.1 moles of water

Explanation:

                     2Al(OH)₃  +   3H₂SO₄   ⇒   Al₂(SO₄)₃  +   6H₂O

                       0.45 mol      0.55 mol                                ?

Process

1.- Calculate the limiting reactant

Theoretical proportion

       Al(OH)₃ / H₂SO₄ = 2/3 = 0.667

Experimental proportion

       Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81

From the proportions, we conclude that the limiting reactant is H₂SO₄

2.- Calculate the moles of H₂O

                        3 moles of H₂SO₄  6 moles of water

                        0.55 moles of H₂SO₄    x

                        x = (0.55 x 6) / 3

                        x = 3.3 / 3

                       x = 1.1 moles of water

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