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

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be