Balance the equation :
Aluminum plus Magnesium iodide yields Magnesium plus Aluminum iodide
(Using organic names and balance chart)

The balanced equation is 2 AlI 3 ( a q ) + 3 Cl 2 ( g ) → 2 AlCl 3 ( a q ) + 3 I 2 ( g ) .

Explanation:

Aluminum has a typical oxidation condition of  3+  , and that of iodine is  1-  .   Along these lines, three iodides can bond with one aluminum. You get  AlI3.  For comparable reasons, aluminum chloride is  AlCl3.  Chlorine and iodine both exist normally as diatomic components, so they are  Cl2(  g  )  also,  I2(  g  ), individually. In spite of the fact that I would anticipate that iodine should be a strong.  

Balancing the equation, we get:  

            2AlI 3(  aq  )  +  3Cl2 (  g  )  →  2AlCl3 (  aq  )   +  3 I 2  (  g  )  

Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the  AlCl  3  on the right.   Normally, presently we have two  Al  on the right, so I multiplied the  AlI  3  on the left. Hence, I have 6  I  on the left, and I needed to significantly increase  I 2  on the right.   We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.

It conforms what ever surface its in.

The answer is !