Answer:
see below
Explanation:
The reaction is a double replacement metathesis reaction that depends upon the formation of a driving force molecule on the product side of the reaction more stable than any on the reactant side of the molecule. Such, in essence, pulls the reaction to completion and prevents reversal.
Driving forces include
1. Insoluble precipitates => need solubility rules for salts in aqueous media
2. Weak Electrolytes (Wk Acids or Weak Bases) => need to review the Arrhenius and Bronsted-Lowery acid-base concepts
3. Gas Decomposition Products => need to understand decomposition of selected compounds gives a gas product that functions as the driving force.
for this problem...
6NaOH(aq) + Al₂(SO₄)₃(aq) => 3Na₂SO₄(aq) + 2Al(OH)₃(s)
The Driving Force of the reaction is Al(OH)₃(s) and prevents rxn reversal by precipitation from the reaction system. The formation of 2Al(OH)₃(s) is feasible as a driving force as all hydroxides except Group IA and IIA metals which are classified as 'strong bases' and prefer to remain in solution 100% ionized. All other OH derivatives will ppt.
FYI => Attached is an abbreviated table of driving forces for metathesis reactions. It is for the most part complete, but always review text references on solubility rules to be sure.
