The balanced equation is 2AlBr₃ + 3Cl₂ ⟶ 2AlCl₃ + 3Br₂
<em>Step 1</em>. Start with the <em>most complicated-looking formula</em> (AlBr₃?) and balance its atoms.
<em>Al</em>: Already balanced — 1 atom each side.
<em>Br</em>: We have 3 Br on the left-hand side and 2 Br on the right. the lowest common multiple of 3 and 2 is 6. Put a 2 in front of AlBr₃ and a 3 in front of Br₂.
2AlBr₃ + Cl₂ ⟶ AlCl₃ + 3Br₂
<em>Step 3</em>: Balance <em>Al</em>
We now need 2Al on the right to balance the 2 Al on the left. Put a 2 in front of AlCl₃.
2AlBr₃ + Cl₂ ⟶ 2AlCl₃ + 3Br₂
<em>Step 4</em>. Balance <em>Cl.
</em>
We have 6 Cl on the right, so we need 6 Cl on the left. Put a 3 in front of Cl₂.
2AlBr₃ + 3Cl₂ ⟶ 2AlCl₃ + 3Br₂
The balanced equation is
2AlBr₃ + 3Cl₂ ⟶ 2AlCl₃ + 3Br₂