Limiting Reactant?

Question

1. Aluminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3 (s) + 3H2SO4 (aq) -----> Al2(SO4)3 (aq) + 6H2O (liquid)

Which reagent is the limiting reactant when 0.500 mol Al(OH)3 and 0.500 mol H2S04 are allowed to react? How many moles of Al2(SO4)3 can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?

Thanks for any help. Reasoning/steps are appreciated :].

Answer 1

The ratio between Al(OH)3 and H2SO4 is 1 : 3
2 : 3 = x : 0.500
x = 0.333 mol Al(OH)3 needed
We have 0.500 mol of Al(OH)3 so it is in excess and H2SO4 is the limiting reactant.
The ratio between H2SO4 ( limiting reactant ) and Al2(SO4)3 is 3 : 1
3 : 1 = 0.500 : x
x = 0.167 mol Al2(SO4)3 produced
0.500 - 0.333 = 0.167 mol Al(OH)3 in excess

Answer 2

From the equation, three moles of H2SO4 are required to react with 2 moles of Al(OH)3 (look at the numbers of the balanced equation - it is necessary to balance these things). So actually 3/2 mole of H2SO4 is necessary to react with 1 mole of Al(OH)3 (3moles of H2SO4 / 2 moles of Al(OH)3 - a mole ratio they talk about). Now if we take the 0.500 moles of Al(OH)3 and multiply by thye mole ratio of 3moles of H2SO4 / 2 moles of Al(OH)3, the moles of Al(OH)3 will cancel and we find that we would need 0.750 moles of H2SO4 to react with the 0.500 moles of Al(OH)3. We don't have 0.750 moles of H2SO4, so we are limited in the reaction to the number of moles of H2SO4 that we have. Hence the H2SO4 is the limiting reagent and we should use it for our calculations since when we run out of it, we are out of the reaction business.

Using the 0.500 moles of H2SO4 as the limiting reagent, and using the mole ratio of 1 mole Al2(SO4)3 / 3 moles of H2SO4 since this puts moles of H2SO4 on the bottom and allows us to cancel it, we find we get (0.500)(1/3) = 0.167 mole of Al2(SO4)3.

Hope this helps

Answer 3

For each mole of Al-hydroxide, 1/2 mole of the Al-sulfate are formed. For each mole of sulfuric acid, 1/3 mole of the Al-sulfate are formed. So the sulfate should be limiting; with equal moles of each reactant, less product is formed based on the acid. By proportion, 1/6th of a mole of the Al-sulfate is formed. By proportion, to form 1/6th of a mole, 1/3 of a mole of the Al-sulfate is needed. So 1/6 of a mole of that compound is unreacted.