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.