# What volume of O_2(g) at 350. "^oC and a pressure of 5.25 atm is needed to completely convert 10.0 g of sulfur to sulfur trioxide?

## Sulfur trioxide, $S {O}_{3}$, is produced in enormous quantities each year for use in the synthesis of sulfuric acid. $S \left(s\right) + {O}_{2} \left(g\right) \to S {O}_{2} \left(g\right)$ $2 S {O}_{2} \left(g\right) + {O}_{2} \left(g\right) \to 2 S {O}_{3} \left(g\right)$

Nov 5, 2016

We can use the stoichiometric equation: $S \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow S {O}_{3} \left(g\right)$. Approx. $5 \cdot L$ dioxygen gas are required.

#### Explanation:

$\text{Moles of sulfur}$ $=$ $\frac{10.0 \cdot g}{32.06 \cdot g \cdot m o {l}^{-} 1} = 0.312 \cdot m o l$.

Given the stoichiometry, we require $\frac{3}{2} \times 0.312 \cdot m o l$ of dioxygen gas, i.e. $0.468 \cdot m o l$.

And now we solve for volume in the Ideal Gas equation:

$V = \frac{n R T}{P}$ $=$ $\frac{0.468 \cdot \cancel{m o l} \times 0.0821 \cdot L \cdot \cancel{a t m} \cdot \cancel{{K}^{-} 1} \cdot \cancel{m o {l}^{-} 1} \times 623 \cdot \cancel{K}}{5.25 \cdot \cancel{a t m}}$ $\cong$ $5 \cdot L$.

Note (i) that we converted $\text{degrees Celsius}$ to $\text{degrees Kelvin}$, and (ii) that we had to use an appropriate $\text{Gas constant, R}$. $\text{R}$ with various units would be supplied as supplementary material in any exam. The fact that our equation gave units of volume, when a volume was sought in the problem, can help to convince us that we got the problem right (for once!!).