# Question #ee990

First we balance the $A l$: 1 in, one out, so that's OK
Then we need 3 $C l$'s at the left (see right hand part:
$A l + 3 H C l \to A l C {l}_{3} + 1 \frac{1}{2} {H}_{2}$ because we have 3 $H$-atoms
$2 A l + 6 H C l \to 2 A l C {l}_{3} + 3 {H}_{2}$