Question

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

Answer 1

Hint:We know that the These reactions also serve to show that aluminum is in fact a more reactive metal than it appears in everyday use and the protective oxide layer of aluminum has to be penetrated by the halogens before the reactions can start, hence the delays, and the need for water to assist the two solid elements getting into contact, in the case of aluminum and iodine keeping this conditions in consideration.

Complete answer:
Aluminum has a common oxidation state of ${{3}^{+}}$ and that of iodine is ${{1}^{-}}$ . So, three iodides can bond with one aluminum. You get $Al{{I}_{3}}$
For similar reasons, aluminum chloride is \[AlC{{I}_{3}}\].
Chlorine and iodine both exist naturally (in their elemental states) as diatomic elements, so they are \[C{{l}_{2}}\left( g \right)\] and \[{{I}_{2}}\left( g \right)\] respectively. Although we would expect iodine to be a solid.
Overall we get:
\[2Al{{I}_{3}}\left( aq \right)+3C{{l}_{2}}\left( g \right)\to 2AlC{{I}_{3}}\left( aq \right)+3{{I}_{2}}\left( g \right)\]
Knowing that there were two chlorines on the left, we just found the common multiple of $2$ and $3$ to be $6$ , and doubled the \[AlC{{I}_{3}}\] on the right.
Naturally, now we have two $Al$ on the right, so I doubled the AlI3 on the left. Thus, we have six $I$ on the left, and I had to triple ${{I}_{2}}$ on the right.

Note:We should keep note that the aluminum iodide is violently reactive in water unless it's a hex hydrate. 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
The clouds of iodine vapour released when aluminum and iodine react can stain the inside of a fume cupboard. Teachers may prefer to demonstrate this reaction outdoors, if possible.