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

Dec 16, 2015

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

For similar reasons, aluminum chloride is ${\text{AlCl}}_{3}$.

Chlorine and iodine both exist naturally (in their elemental states) as diatomic elements, so they are ${\text{Cl}}_{2} \left(g\right)$ and ${\text{I}}_{2} \left(g\right)$, respectively. Although I would expect iodine to be a solid...

Overall we get:

$2 {\text{AlI"_3(aq) + 3"Cl"_2(g) -> 2"AlCl"_3(aq) + 3"I}}_{2} \left(g\right)$

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 ${\text{AlCl}}_{3}$ on the right.

Naturally, now we have two $\text{Al}$ on the right, so I doubled the ${\text{AlI}}_{3}$ on the left. Thus, I have 6 $\text{I}$ on the left, and I had to triple ${\text{I}}_{2}$ 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.