Aluminum sulfate, $A l_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3$ , decomposes to form aluminum oxide, $A l_{2} O_{3}$ $Al_2O_3$ , and sulfur trioxide, $S O_{3}$ $SO_3$ . How do you write the balanced equation for this reaction?

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Answer 1 · 2016-01-08T14:11:23

$A l_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3$ $\to$ $->$ $A l_{2} O_{3}$ $Al_2O_3$ + $3 S O_{3}$ $3SO_3$

Putting 3 as cooefficient for sulfur trioxide will balance the chemical equation. Check the number of atoms of each element by having an inventory.

Reactant Side:
$A l = 2 \times 1 = 2$ $Al=2xx1=2$
$S = 3 \times 1 \times 1 = 3$ $S=3xx1xx1=3$
$O = 3 \times 4 \times 1 = 12$ $O=3xx4xx1=12$

Product Side:
$A l = 2 \times 1 = 2$ $Al=2xx1=2$
$S = 1 \times 3 = 3$ $S=1xx3=3$
$O = 3 \times 1 + 3 \times 3 = 12$ $O=3xx1 + 3xx3 = 12$

Aluminum sulfate, Al2(SO4)3Al_2(SO_4)_3, decomposes to form aluminum oxide, Al2O3Al_2O_3, and sulfur trioxide, SO3SO_3. How do you write the balanced equation for this reaction?