#S_8(s) + 12O_2(g) rarr 8SO_3(g)#; (i)

Alternatively,

#1/8S_8(s) + 3/2O_2(g) rarr SO_3(g)# (ii)

Note that this gives the stoichiometric conversion of #S(0)# to #S(VI)#. Industrially sulfur trioxide (and what do you think this is used for?) derives from #SO_2(g)#.

i.e. #1/8S_8(s) +O_2(g) rarr SO_2(g)#; (iii)

then #SO_2(g) + 1/2O_2(g) rarr SO_3(g)# (iv)

In the third equation I used the #1/8# coefficient to cope with the molecularity of the sulfur ring (then I used a #1/2# coefficient in the fourth equation; I am certainly free to do this). I am not free to leave the equation unbalanced. Is it balanced? Don't trust my arithmetic. The fourth equation to give #SO_3# uses (I think) a supported #V_2O_5# catalyst of some sort. It must be an incredibly dirty and smelly process, yet it is undoubtedly vital to our civilization.

Note that #SO_3(g)# is often added to sulfuric acid to give #H_2S_2O_7#, pyrosulfuric acid.