Call

Aluminum has a common oxidation state of $"3+"$, and that of iodine is $"1-"$. So, three iodides can bond with one aluminum. You get $"AlI"_3$.

For similar reasons, aluminum chloride is $"AlCl"_3$.

Chlorine and iodine both exist naturally (in their elemental states) as diatomic , so they are $"Cl"_2(g)$ and $"I"_2(g)$, respectively. Although I would expect iodine to be a solid...

Overall we get:

$2"AlI"_3(aq) + 3"Cl"_2(g) -> 2"AlCl"_3(aq) + 3"I"_2(g)$

Knowing that there were two chlorines on the left, I just found the common multiple of 2 and 3 to be 6, and doubled the $"AlCl"_3$ on the right.

Naturally, now we have two $"Al"$ on the right, so I doubled the $"AlI"_3$ on the left. Thus, I have 6 $"I"$ on the left, and I had to triple $"I"_2$ on the right.

We should note, though, that aluminum iodide is violently reactive in water unless it's a hexahydrate. So, it's probably the anhydrous version dissolved in water, and the amount of heat produced might explain why iodine is a gaseous product, and not a solid.