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

Apr 15, 2016

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

$3 B a C {l}_{2} + A {l}_{2} {S}_{3} \to 3 B a S + 2 A l C {l}_{3}$

#### Explanation:

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

$a B a C {l}_{2} + b A {l}_{2} {S}_{3} \to c B a S + \mathrm{dA} l C {l}_{3}$

For each element, we can have a balance equation:

$B a : a = c$

$C l : 2 a = 3 d$

$A l : 2 b = d$

$S : 3 b = c$

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

$a = 1$

$c = a = 1$

$3 d = 2 a \iff d = \frac{2}{3}$

$2 b = d \iff b = \frac{\frac{2}{3}}{2} = \frac{1}{3}$

Now the equation can be balanced:

$B a C {l}_{2} + \frac{1}{3} A {l}_{2} {S}_{3} \to B a S + \frac{2}{3} A l C {l}_{3}$

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:

$3 B a C {l}_{2} + A {l}_{2} {S}_{3} \to 3 B a S + 2 A l C {l}_{3}$