How would you balance the following equation: S8 + O2 -- SO3?

Question

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

Jeremiah Adams · Accepted Answer

With ease! Or at least it's easy if you have a chemical industry.

#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.

Owen Ambrose · Answer

undefined To balance the equation S8 + O2 -> SO3, you would need to multiply the oxygen molecule (O2) by 3 and the sulfur dioxide (SO3) by 1. The balanced equation would be:

8S8 + 12O2 -> 8SO3

How would you balance the following equation: S8 + O2 --&gt; SO3?

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