How would you balance the following equation: aluminum iodide and chlorine gas react to form aluminum chloride and iodine gas?

1 Answer
Dec 16, 2015

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 elements, 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.