# How do you balance S_8 + O_2 -> SO_3?

Jan 21, 2016

Well I could do it directly,

$\frac{1}{8} {S}_{8} \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow S {O}_{3} \left(g\right)$

#### Explanation:

But because some might disagree with the use of a non-integral stoichiometric coefficient, we could use integral coefficients:

${S}_{8} \left(s\right) + 12 {O}_{2} \left(g\right) \rightarrow 8 S {O}_{3} \left(g\right)$

Both reactions are formalisms. They do represent observed stoichiometry, but they are not very useful in giving us more chemical insight.

Industrially, this reaction is performed by the oxidation of $S {O}_{2} \left(g\right)$ by oxygen in the presence of some form of solid catalyst.

$S {O}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow S {O}_{3} \left(g\right)$

All of these schemes are representations only of actual chemical reactions. They must be incredibly dirty and smelly processes, yet they are undoubtedly vital to our industrial society. Sulfur trioxide is the acid anhydride of sulfuric acid, ${H}_{2} S {O}_{4}$.