What Is The Theoretical Yield Of So3, Solved Produced By The
The question specifies that we got #1.88xx10^23# #"dioxygen molecules"#...i.e. a molar quantity of...
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#(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....
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#n("SO"_3) =0.208 color(white)(l) mol#
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#
#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.
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