Answer
Balanced equation:
$Al_2(SO_4)_3(aq) + 6NaOH(aq) --\gt 3Na_2SO_4(aq) + 2Al(OH)_3(s)$
Reaction type: Double replacement.
Work Step by Step
$Al_2(SO_4)_3(aq) + NaOH(aq) --\gt Na_2SO_4(aq) + Al(OH)_3(s)$
1. Balance the number of sulfur atoms, by putting a "3" in front of $Na_2SO_4$.
$Al_2(SO_4)_3(aq) + NaOH(aq) --\gt 3Na_2SO_4(aq) + Al(OH)_3(s)$
2. Balance the number of sodium atoms, by putting a "6" in front of $NaOH$
$Al_2(SO_4)_3(aq) + 6NaOH(aq) --\gt 3Na_2SO_4(aq) + Al(OH)_3(s)$
3. Balance the number of aluminum atoms, by putting a "2" in front of $Al(OH)_3$.
$Al_2(SO_4)_3(aq) + 6NaOH(aq) --\gt 3Na_2SO_4(aq) + 2Al(OH)_3(s)$
4. Classify the reaction.
That reaction follows the pattern for a double replacement reaction (AB + CD $--\gt$ AD + BC).
