Elemental S reacts with O2 to form SO3 according to the reaction 2S+3O2→2SO3 Part B: What is the theoretical yield of SO3 produced by the quantities described in Part A? Express your answer numerically in grams.

The question specifies that we got #1.88xx10^23# #"dioxygen molecules"#...i.e. a molar quantity of...

#(1.88xx10^23*"molecules")/(6.022xx10^23*"molecules"*mol^-1)=0.312*mol#...

But we gots with respect to sulfur, #(6.67*g)/(32.06*g*mol^-1)=0.208*mol#...

And a bit of arithmetic later, we establish that we got stoichiometric quantities of dioxygen, and sulfur….in the reaction we produce a mass of ………..

#0.208*molxx80.07*g*mol^-1=16.65*g#.

Note that when #"sulfur trioxide"# is made industrially (and this a very important commodity chemical), sulfur is oxidized to #SO_2#, and this is then oxidized up to #SO_3# with some catalysis...

#SO_2(g) + 1/2O_2(g) stackrel(V_2O_5)rarrSO_3(g)#

#SO_3(g) + H_2O(l) rarr underbrace(H_2SO_4(aq))_"sulfuric acid"#

The industrial sulfur cycle must be a dirty, smelly, unfriendly process. The process is undoubtedly necessary to support our civilization....

Numerical relationships between coefficients of the equation suggest the following conversion factors:

• #(n("SO"_3))/(n("O"_2)) = 2/3#
• #(n("SO"_3))/(n("S")) = 2/2 = 1#

From part A:

#n("O"_2) = (N("O"_2))/(N_A)#

Where #n("O"_2)# the number of moles of oxygen molecules available (in #mol#), #N("O"_2)# the corresponding number of oxygen molecules available (a large, typically unitless number), and #N_A# the Avogadro's number.

#n("O"_2) = (1.88 xx 10^(23)) / (6.023 xx 10^(23) color(white)(l) mol^(-1)) = 0.312 color(white)(l) mol#

Apply the conversion factor between #n("SO"_3)# and #n("O"_2)#,

#n("SO"_3) = (n("SO"_3))/(n("O"_2)) * n("O"_2) #
#color(white)(n("SO"_3)) = 2/3 * 0.312 color(white)(l) mol#
#color(white)(n("SO"_3)) = 0.208 color(white)(l) mol#

As a side note: in cases that the formula mass of sulfur, #"S"#, is available, calculating #n("S")# from #m("S")# and #M("S")# and thus #n("SO"_3)# given #n("SO"_3)//n("S") = 1# is expected to yield the same result.