# Question #5db3b

May 2, 2016

Approx. $15 \cdot g$

#### Explanation:

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

$\text{Moles of sulfur dioxide}$ $=$ $\frac{12.4 \cdot g}{64.06 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.194 \cdot m o l$.

$\text{Moles of dioxygen}$ $=$ $\frac{3.45 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.108 \cdot m o l$.

Clearly, from the molar quantities there is a slight excess of oxygen. And since, to start, there are $0.194 \cdot m o l$ $S {O}_{2}$, to finish there would be an equimolar quantity of $S {O}_{3} \left(g\right)$ given quantitative yield.

$\text{Mass of } S {O}_{3}$ $=$ $0.194 \cdot m o l \times 80.07 \cdot g \cdot m o {l}^{-} 1 = 15.5 \cdot g$