# What volume of O_2 (g) at 350^oC and a pressure of 5.25 atm is needed to completely convert 5.00 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. $S \left(s\right) + {O}_{2} \left(g\right) \to S {O}_{2} \left(g\right)$ $2 S {O}_{2} \left(g\right) + O 2 \left(g\right) \to 2 S {O}_{3} \left(g\right)$

Jun 22, 2017

Approx.....$2 \cdot L$

#### Explanation:

We can combine the given equations........

$S \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow S {O}_{3} \left(g\right)$

And this clearly gives us the mass transfer; 1 equiv of sulfur is oxidized by 3/2 equiv dioxygen gas.....

$\text{Moles of sulfur} = \frac{5.00 \cdot g}{32.06 \cdot g \cdot m o {l}^{-} 1} = 0.156 \cdot m o l$

And so we need $0.156 \cdot m o l \times \frac{3}{2} \times 32.00 \cdot g \cdot m o {l}^{-} 1 = 7.49 \cdot g$ with respect to dioxygen gas......

And we use the Ideal Gas Equation.......

$V = \frac{n R T}{P} = \frac{\frac{7.49 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1} \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 623.2 \cdot K}{5.25 \cdot a t m}$

=??L