Question

al+ i₂ → ali₃

reactants	inventory	products
	i

Explanation:

Step1: Balance aluminum (Al)

On the product side, there is 1 Al in $AlI_3$. So, we put a 2 in front of Al on the reactant side to balance Al atoms. The equation becomes $2Al + I_2
ightarrow 2AlI_3$.

Step2: Balance iodine (I)

On the product side, there are 6 I atoms in $2AlI_3$. On the reactant side, $I_2$ contains 2 I atoms. So, we put a 3 in front of $I_2$ to balance I atoms. The balanced equation is $2Al+3I_2
ightarrow 2AlI_3$.

Answer:

2; 3; 2