How do you balance $B a C l_{2} + A l_{2} S_{3} \to B a S + A l C l_{3}$ $BaCl_2 + Al_2S_3 -> BaS + AlCl_3$ ?

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How do you balance $B a C l_{2} + A l_{2} S_{3} \to B a S + A l C l_{3}$ $BaCl_2 + Al_2S_3 -> BaS + AlCl_3$ ?

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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:

$3 B a C l_{2} + A l_{2} S_{3} \to 3 B a S + 2 A l C l_{3}$ $3BaCl_2+Al_2S_3->3BaS+2AlCl_3$

Explanation:

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

$a B a C l_{2} + b A l_{2} S_{3} \to c B a S + d A l C l_{3}$ $aBaCl_2+bAl_2S_3->cBaS+dAlCl_3$

For each element, we can have a balance equation:

$B a : a = c$ $Ba: a=c$

$C l : 2 a = 3 d$ $Cl: 2a=3d$

$A l : 2 b = d$ $Al: 2b=d$

$S : 3 b = c$ $S: 3b=c$

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

$a = 1$ $a=1$

$c = a = 1$ $c=a=1$

$3 d = 2 a \Leftrightarrow d = \frac{2}{3}$ $3d=2a<=>d=2/3$

$2 b = d \Leftrightarrow b = \frac{\frac{2}{3}}{2} = \frac{1}{3}$ $2b=d<=>b=(2/3)/2=1/3$

Now the equation can be balanced:

$B a C l_{2} + \frac{1}{3} A l_{2} S_{3} \to B a S + \frac{2}{3} A l C l_{3}$ $BaCl_2+1/3Al_2S_3->BaS+2/3AlCl_3$

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:

$3 B a C l_{2} + A l_{2} S_{3} \to 3 B a S + 2 A l C l_{3}$ $3BaCl_2+Al_2S_3->3BaS+2AlCl_3$

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How do you balance $B a C l_{2} + A l_{2} S_{3} \to B a S + A l C l_{3}$ $BaCl_2 + Al_2S_3 -> BaS + AlCl_3$ ?

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Explanation:

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How do you balance BaCl2+Al2S3→BaS+AlCl3BaCl_2 + Al_2S_3 -> BaS + AlCl_3?

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Explanation:

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How do you balance $B a C l_{2} + A l_{2} S_{3} \to B a S + A l C l_{3}$ $BaCl_2 + Al_2S_3 -> BaS + AlCl_3$ ?