# 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.

Jul 16, 2018

We address the equation...$S \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow S {O}_{3} \left(g\right)$

#### Explanation:

The question specifies that we got $1.88 \times {10}^{23}$ $\text{dioxygen molecules}$...i.e. a molar quantity of...

$\left(1.88 \times {10}^{23} \cdot \text{molecules")/(6.022xx10^23*"molecules} \cdot m o {l}^{-} 1\right) = 0.312 \cdot m o l$...

But we gots with respect to sulfur, $\frac{6.67 \cdot g}{32.06 \cdot g \cdot m o {l}^{-} 1} = 0.208 \cdot m o l$...

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 ………..

$0.208 \cdot m o l \times 80.07 \cdot g \cdot m o {l}^{-} 1 = 16.65 \cdot g$.

Note that when $\text{sulfur trioxide}$ is made industrially (and this a very important commodity chemical), sulfur is oxidized to $S {O}_{2}$, and this is then oxidized up to $S {O}_{3}$ with some catalysis...

$S {O}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \stackrel{{V}_{2} {O}_{5}}{\rightarrow} S {O}_{3} \left(g\right)$

$S {O}_{3} \left(g\right) + {H}_{2} O \left(l\right) \rightarrow {\underbrace{{H}_{2} S {O}_{4} \left(a q\right)}}_{\text{sulfuric acid}}$

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

Jul 16, 2018

$n \left({\text{SO}}_{3}\right) = 0.208 \textcolor{w h i t e}{l} m o l$

#### Explanation:

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

• $\left(n \left({\text{SO"_3))/(n("O}}_{2}\right)\right) = \frac{2}{3}$
• $\left(n \left(\text{SO"_3))/(n("S}\right)\right) = \frac{2}{2} = 1$

From part A:

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

Where $n \left({\text{O}}_{2}\right)$ the number of moles of oxygen molecules available (in $m o l$), $N \left({\text{O}}_{2}\right)$ the corresponding number of oxygen molecules available (a large, typically unitless number), and ${N}_{A}$ the Avogadro's number.

$n \left({\text{O}}_{2}\right) = \frac{1.88 \times {10}^{23}}{6.023 \times {10}^{23} \textcolor{w h i t e}{l} m o {l}^{- 1}} = 0.312 \textcolor{w h i t e}{l} m o l$

Apply the conversion factor between $n \left({\text{SO}}_{3}\right)$ and $n \left({\text{O}}_{2}\right)$,

$n \left({\text{SO"_3) = (n("SO"_3))/(n("O"_2)) * n("O}}_{2}\right)$
$\textcolor{w h i t e}{n \left({\text{SO}}_{3}\right)} = \frac{2}{3} \cdot 0.312 \textcolor{w h i t e}{l} m o l$
$\textcolor{w h i t e}{n \left({\text{SO}}_{3}\right)} = 0.208 \textcolor{w h i t e}{l} m o l$

As a side note: in cases that the formula mass of sulfur, $\text{S}$, is available, calculating $n \left(\text{S}\right)$ from $m \left(\text{S}\right)$ and $M \left(\text{S}\right)$ and thus $n \left({\text{SO}}_{3}\right)$ given $n \left(\text{SO"_3)//n("S}\right) = 1$ is expected to yield the same result.