Question

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?

Answer 1

The answer to your question is 1.1 moles of water

Step-by-step 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