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?

a. 0.183
b. 0.367 mol
c. 1.10 mol
d. 0.250 mol
e. 0.125 mol

Option c. 1.10 moles

Explanation:

By the moles of the two reactants, we need to determine the limiting reactant.

If we only have the mass (or the moles) of one, we assume the other is in excess.

The reaction is: 2Al(OH)₃(s) + 3H₂SO₄(aq)  → Al₂(SO₄)₃(aq) + 6H₂O(l)

Let's work with stoichiometry: 2 moles of hidroxide react with 3 moles of sulfuric acid.

Then, 0.450 moles of hidroxide, will react with (0.450 . 3) / 2 = 0.675 moles of acid. Notice we have 0.550 of it, so we do not have enough acid to react, therefore the H₂SO₄ is the limiting reagent

Let's do another rule of three. Ratio is 3:6

3 moles of acid can produce 6 moles of water

Then, 0.550 moles of acid, will produce (0.550 . 6) / 3 = 1.1 moles of water

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa