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.

Part A: 1.88x10^23 O2 molecules are needed to react with 6.67 g of S.

2 Answers
Jul 16, 2018

We address the equation...S(s) + 3/2O_2(g) rarr SO_3(g)

Explanation:

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....

Jul 16, 2018

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

Explanation:

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.