Question

Aluminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3(s)+3H2SO4(aq)→Al2(SO4)3(aq)+6H2O(l)

1)How many moles of Al2(SO4)3 can form under these conditions?

2)How many moles of the excess reactant remain after the completion of the reaction?

Thank you!!

Answer 1

We have to find limiting reactant

Given moles of

Both the reactant = 0.700 mol

We calculate moles of Al2(SO4)3 by using both the given moles

Moles of Al2(SO4)3 from moles of Al(OH)3

Mole ratio of Al2(SO4)3 and Al(OH)3 is 1 : 2

Mols of Al2(SO4)3 = moles of Al(OH)2 * 1 mol Al2(SO4)3 / 2 mol Al(OH)3

= 0.700 mol Al(OH)3 * 1 mol Al2(SO4)3 / 2 mol Al(OH)3

= 0.35 mol Al2(SO4)3

Now moles of Al2(SO4)3 from moles of H2SO4

Mol ratio of Al2(SO4)3 to H2SO4 is 1 : 3

Mol of Al2(SO4)3 = mols of H2SO4 *1 mol of Al2(SO4)3 / 3 mol H2SO4

= 0.700 mol H2SO4 * 1 mol Al2(SO4)3 / 3 mol H2SO4

= 0.233 mol Al2(SO4)3

Moles of H2SO4 gives less number of moles of Al2(SO4)3 and hence H2SO4 is limiting reactant

So only limiting reactant limits the product formation so mol of Al2(SO4)3 formed

From limiting reactant would be maximum.

Moles of Al2(SO4)3 = 0.233 mol

Question 2 )

Excess reactant is 2 Al(OH)3

We find moles of Al(OH)3 required to react with H2SO4 completely

Mol ratio of H2SO4 : Al(OH)3 is 3 : 2

Mol of Al(OH)3 = mol of H2SO4 * 2 mol Al(OH)3 / 3 mol H2SO4

= 0.700 mol H2SO4 * 2 mol Al(OH)3 / 3 mol H2SO4

= 0.467 mol Al(OH)3

Excess reactant remains = Original moles – reacted moles

= 0.700 -0.467

= 0.233 moles of reactant will remains at the completion of the reaction