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

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Answer 1 · 2018-05-09T07:50:14

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

$S_{8} (s) + 12 O_{2} (g) \to 8 S O_{3}$ $S_8(s) +12O_2(g) rarr 8SO_3$

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

$\frac{1}{8} S_{8} (s) + \frac{3}{2} O_{2} \to S O_{3} (g)$ $1/8S_8(s) + 3/2O_2 rarr SO_3(g)$

Industrially, this is performed by the $contact process$ $"contact process"$ . Sulfur is burned in air...

$\frac{1}{8} S_{8} (g) + O_{2} (g) \to S O_{2} (g)$ $1/8S_8(g) + O_2(g) rarr SO_2(g)$

And then this is further oxidized to $S O_{3}$ $SO_3$ by means of a vanadium pentoxide catalyst...

$S O_{2} (g) + \frac{1}{2} O_{2} (g) V_{2} O_{5} - - \to S O_{3} (g)$ $SO_2(g) +1/2O_2(g)stackrel(V_2O_5)rarrSO_3(g)$

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