How to work out equations with period 3 oxides?
The first thing to say is that you final equation doesn't balance. In fact it is impossible to balance your equation using hydroxide because you can never conserve charge with only neutral products.

You can write a reaction that forms $\ce{Al(OH)3}$: $$\ce{Al2O3 + 3H2O -&gt; 2Al(OH)3}$$

However, beyond about pH 6 the $\ce{Al(OH)3}$ will start to dissolve to form $\ce{[Al(OH)4]-}$. Once you get to around pH 7.5 all of the solid has essentially dissolved and you are left with solution of $\ce{Na+}$ and $\ce{[Al(OH)4]-}$ (see footnote). So unless you are very sparing with your addition on $\ce{NaOH}$ you will get the tetrahydroxo product in solution.

$$\ce{Al(OH)3 + OH- -&gt; [Al(OH)4]-}$$

Footnote:
The aluminium species present in aqueous solution are strongly dependent on the pH, particularly around the neutral point. There is a nice graph here showing the different species. The notation is $\ce{Al}$ corresponds to $\ce{[Al(H2O)6]^{3+}}$ and then $\ce{AlH_{-n}}$ denotes the loss of n protons from the hexaaqua complex. This was originally found by @Jan for this question.