# How would you balance: S8 + O2 --> SO3?

May 9, 2018

Stoichiometrically...garbage in equals garbage out....

#### Explanation:

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

To make this a little easier we could use fractional coefficients...

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

Industrially, this is performed by the $\text{contact process}$. Sulfur is burned in air...

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

And then this is further oxidized to $S {O}_{3}$ by means of a vanadium pentoxide catalyst...

$S {O}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \stackrel{{V}_{2} {O}_{5}}{\rightarrow} S {O}_{3} \left(g\right)$

This must be a very dirty, nasty, and smelly process. It is undoubtedly necessary to perform this reaction to produce an industrial feedstock.