Elemental S reacts with O2 to form SO3 according to the reaction 2S+3O2→2SO3 Part A: 1.88x10^23 O2 molecules are needed to react with 6.67 g of S Part B: What is the theoretical yield of SO3 produced by the quantities described in Part A? Express in grams

1 Answer
Mar 11, 2018

Well, we address the stoichiometric equation...

#S(s) +3/2O_2(g) rarr SO_3(g)#

Explanation:

#"Moles of sulfur"=(6.67*g)/(32.06*g*mol^-1)=0.208*mol#

And #"Moles of dioxygen"=(1.88xx10^23*"molecules")/(6.022xx10^23*"molecules"*mol^-1)=#

#=0.312*mol#.

But you have not told us how much sulfur trioxide you get, so we cannot address the yield.

Certainly you have stoichiometric quantities of sulfur and dioxygen. The industrial preparation of sulfur trioxide takes sulfur dioxide, and combines this with dioxygen over a catalytic surface...

#SO_2(g) +1/2O_2(g) stackrel("catalysis")rarrSO_3(g)#