# How many liters of O_2 gas, measured at 777 mm Hg and 35°C, are required to completely react with 2.7 mol of Al?

Nov 19, 2016

Approx. $50 \cdot L$ of $\text{dioxygen gas}$

#### Explanation:

A measurement of $777 \text{ mm Hg}$ is completely unrealistic. Given $1 \cdot a t m \equiv 760 \cdot m m \cdot H g$, a column of mercury that is longer than $760 \cdot m m$ risks getting mercury all over the shop, and this is an outcome that you should avoid.

Given $777 \text{ mm Hg}$ $=$ $\frac{777 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} = 1.02 \cdot a t m ,$ we need a stoichiometric equation:

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

So if there are $2.7 \cdot m o l$ of metal, we need $2.025 \cdot m o l$ of $\text{dioxygen gas}$.

$V = \frac{n R T}{P}$ $=$ $\frac{2.025 \cdot m o l \times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 308 \cdot K}{1.02 \cdot a t m}$ =??*L