al+ i₂ → ali₃\nreactants inventory products\nal\ni\nremember: coefficients are the numbers in front of each…

Explanation:

Step1: Balance the iodine atoms

On the right - hand side, there are 3 iodine atoms in $AlI_3$. On the left - hand side, iodine is in the form of $I_2$. To balance the iodine, we need to find the least common multiple of 2 and 3, which is 6. So we put a 3 in front of $I_2$ and a 2 in front of $AlI_3$. $Al + 3I_2\rightarrow2AlI_3$

Step2: Balance the aluminum atoms

Now, since there are 2 aluminum atoms in $2AlI_3$, we put a 2 in front of $Al$ on the left - hand side. The balanced equation is $2Al+3I_2\rightarrow2AlI_3$. The coefficient ratio of $Al:I_2:AlI_3$ is $2:3:2$.

Answer:

$2:3:2$