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

1 Answer
Apr 15, 2016

Answer:

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#