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

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

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Answer 1 · 2015-12-16T02:26:05

Aluminum has a common oxidation state of $3+$ $"3+"$ , and that of iodine is $1-$ $"1-"$ . So, three iodides can bond with one aluminum. You get ${AlI}_{3}$ $"AlI"_3$ .

For similar reasons, aluminum chloride is ${AlCl}_{3}$ $"AlCl"_3$ .

Chlorine and iodine both exist naturally (in their elemental states) as diatomic elements, so they are ${Cl}_{2} (g)$ $"Cl"_2(g)$ and $I_{2} (g)$ $"I"_2(g)$ , respectively. Although I would expect iodine to be a solid...

Overall we get:

$2 {AlI}_{3} (a q) + 3 {Cl}_{2} (g) \to 2 {AlCl}_{3} (a q) + 3 I_{2} (g)$ $2"AlI"_3(aq) + 3"Cl"_2(g) -> 2"AlCl"_3(aq) + 3"I"_2(g)$

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 ${AlCl}_{3}$ $"AlCl"_3$ on the right.

Naturally, now we have two $Al$ $"Al"$ on the right, so I doubled the ${AlI}_{3}$ $"AlI"_3$ on the left. Thus, I have 6 $I$ $"I"$ on the left, and I had to triple $I_{2}$ $"I"_2$ 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.

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

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