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

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

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Answer 1 · 2016-04-15T13:23:36

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:

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

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

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How do you balance BaCl_2 + Al_2S_3 -> BaS + AlCl_3BaCl_2 + Al_2S_3 -> BaS + AlCl_3?

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