The reaction here just seems to be a double-replacement (now called "salt metathesis" apparently). I can tell this because we have two metals, and each metal is bonded to a different molecule. In these cases, the metals always exchange their buddy nonmetals or polyatomic ions with the other metal in the reaction. We're going to ignore the water for now, so we can concentrate on getting the right reaction. The water is simply complexed to the Al2O3, and doesn't participate in any covalent bonding...yet...

Now, remember, these particular metals only have one stable valence when bonding with other elements: Al becomes Al3+ and Na becomes Na+. Using that fact, we can determine (if you weren't aware already) that the charge on CO3 is 2-. O always is 2- when bonded in molecules like this.

Now, we need to think about how the double replacement reaction will proceed.

First, the CO3 will separate from the Na and bond with the aluminum to produce some combination of Al and CO3. Considering that Al will have a valence of 3+ and CO3 has a valence of 2-, we simply have to find the number combo that will match the positives and negatives (if you're into math, that means find the least common multiple). In this case, we need 2 aluminums and 3 carbonates (CO3) to make Al2(CO3)3.

Now, we know Na needs to bond with the oxygen that the aluminum is becoming detached from. Because Na has valence +1 and O has valence 2-, we know the formula must be Na2O.

Our equation has become the following:

Al2O3*2H20 + 3Na2CO3 -> Al2(CO3)3 + 3Na2O + H20

Now, let's go back and talk about the water on the aluminum oxide at the beginning.

Na2O (sodium oxide) and Al2(CO3)3 react with water to make pretty common chemicals:

Na2O + H2O -> NaOH (sodium hydroxide)

Al2(CO3)3 + H20 -> Al(OH)3 + CO2 (Aluminum Hydroxide + carbon dioxide)

Therefore, our final reaction, taking into account that we're dealing with water, becomes:

Al2O3*2H2O + Na2CO3 -> NaOH + Al(OH)3 + CO2

Hope that helps!