Question

How do you balance `BaCl_2 + Al_2S_3 -> BaS + AlCl_3`?

Answer 1

Create an equation for each of the elements, then set one of them and solve for the others. Answer is:

`3BaCl_2+Al_2S_3->3BaS+2AlCl_3`

Explanation:

Let the four balancing factors be `a` `b` `c` `d` as follows:

`aBaCl_2+bAl_2S_3->cBaS+dAlCl_3`

For each element, we can have a balance equation:

`Ba: a=c`

`Cl: 2a=3d`

`Al: 2b=d`

`S: 3b=c`

You can notice that if you set one of these factors it will chain you to the next factor. Let's set `a=1`

`a=1`

`c=a=1`

`3d=2a<=>d=2/3`

`2b=d<=>b=(2/3)/2=1/3`

Now the equation can be balanced:

`BaCl_2+1/3Al_2S_3->BaS+2/3AlCl_3`

However, many don't like the fractions because of the molecule concept. Since they have a common denuminator, multiply everything by 3 to get ride of the fractions:

`3BaCl_2+Al_2S_3->3BaS+2AlCl_3`