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.
