Na2CO3 + FeBr3 + H2O: why do we get Fe(OH)3?
You have certainly already recognized $\ce{Na2CO3}$ as the base in your equation.

Did you also consider the following:
When $\ce{FeBr3}$ is dissolved in water, the hexaaqua complex $\ce{[Fe(H2O)6]^3+}$ is formed. This complex is fairly acidic:
$$\ce{[Fe(H2O)6]^3+ + H2O&lt;=&gt; [Fe(H2O)5OH]^2+ + H3O+}$$

Ignoring ligand exchange processes, you might state that the $\ce{OH}$ groups come from the deprotonation of the water ligands of your initial complex.