L __Al + ___HCI → __AlCl3 + __H2

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Al + HCl → AlCl₃ + H₂

Balance Cl.

1 on the left, 3 on the right. Add a coefficient of 3 in front of HCl.

Al + 3HCl → AlCl₃ + H₂

Balance H.

3 on the left, 2 on the right. We have to start by multiplying everything else by 2.

2Al + 3HCl → 2AlCl₃ + 2H₂

Now we have 2 on the right and 4 on the left. Change the coefficient in front of HCl from 3 to 4.

2Al + 4HCl → 2AlCl₃ + 2H₂

Now, for Cl, we have 4 on the left and 6 on the right. Change the coefficient in front of HCl again from 4 to 6.

2Al + 6HCl → 2AlCl₃ + 2H₂

Now, our H is unbalanced again. 6 on the left, 4 on the right. Change the coefficient in front of H₂ from 2 to 3.

2Al + 6HCl → 2AlCl₃ + 3H₂

Balance Al.

2 on the left, 2 on the right. Already balanced.

Here is our final balanced equation:

2Al + 6HCl → 2AlCl₃ + 3H₂

Hope this helps!

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