# What is the total pressure (mmHg) of a gaseous mixture of 3.7 g of hydrogen gas and 9.1 g of oxygen gas in a 3.24-L flask if the partial pressure of hydrogen is 318 mmHg?

Nov 25, 2016

Approx. $1.5 \cdot a t m$.

#### Explanation:

Given constant temperature, and volume, which we certainly have here (even though we don't know their magnitudes), pressure is proportional to the number of moles of gas, or vice versa.

$\text{Moles of dihydrogen}$ $=$ $\frac{3.7 \cdot g}{2.02 \cdot g \cdot m o {l}^{-} 1} = 1.83 \cdot m o l$.

$\text{Moles of dioxygen}$ $=$ $\frac{9.1 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1} = 4.97 \cdot m o l$.

We are given that ${P}_{\text{dihydrogen}} = 318 \cdot m m \cdot H g = \frac{318 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} = 0.418 \cdot a t m$

And given the proportionality between moles and pressure,

${P}_{\text{dioxygen}} = \frac{4.97 \cdot m o l}{1.83 \cdot m o l} \times 0.418 \cdot a t m = 1.135 \cdot a t m$.

And finally, ${P}_{\text{Total"=P_"dihydrogen"+P_"dioxygen}}$

$=$ $\left(0.418 + 1.135\right) \cdot a t m$

Note that I converted the $m m \cdot H g$ to units of $\text{atmospheres}$, because I really don't think $m m \cdot H g$ should be used for pressures much above $1 \cdot a t m , i . e . 760 \cdot m m \cdot H g$. There seems to be quite a few questions on these boards where mercury columns greater than $760 \cdot m m \cdot H g$ have been used to measure pressure. Obviously, the person who set the question has never had to clean up a mercury spill.