The question specifies the we obtained #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 come sulfur, #(6.67*g)/(32.06*g*mol^-1)=0.208*mol#...

And a little bit of arithmetic later, we develop that we acquired stoichiometric amounts of dioxygen, and also sulfur….in the reaction we produce a mass of ………..

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

Note that once #"sulfur trioxide"# is do industrially (and this a an extremely important commodity chemical), sulfur is oxidized to #SO_2#, and also this is climate oxidized approximately #SO_3# v 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 commercial sulfur cycle have to be a dirty, smelly, unfriendly process. The process is undoubtedly vital to support our civilization....

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Jacob T.
Jul 16, 2018

#n("SO"_3) =0.208 color(white)(l) mol#

Explanation:

Numerical relationships in between coefficients the the equation imply the adhering to 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 variety of moles of oxygen molecules obtainable (in #mol#), #N("O"_2)# the corresponding variety of oxygen molecules easily accessible (a large, frequently unitless number), and also #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 instances 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 exact same result.