Question

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.

Answer 1

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....

Answer 2

`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.