Question

Question #ee990

Answer 1

First we balance the `Al`: 1 in, one out, so that's OK
Then we need 3 `Cl`'s at the left (see right hand part:

`Al+3HCl->AlCl_3+1 1/2 H_2` because we have 3 `H`-atoms

We can't have half molecules so we multiply everything by 2:

`2Al+6HCl->2AlCl_3+3H_2`