# What volume of O_2 gas (in L), measured at 781 mmHg and 34 degrees C, is required to completely react with 53.1 g of Al?

Dec 15, 2016

Again, a mercury column has been used to measure a pressure over one atmosphere. This is not something that should be attempted. I get a volume of approx. $70 \cdot L$.

#### Explanation:

We need a stoichiometrically balanced equation:

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

$\text{Moles of aluminum metal}$ $=$ $\frac{53.1 \cdot g}{27.0 \cdot g \cdot m o {l}^{-} 1} = 1.97 \cdot m o l$

Thus, we need $1.97 \cdot m o l \times \frac{3}{2} \cdot m o l$ $\text{dioxygen gas}$ $=$ $2.95 \cdot m o l$.

And now, we simply use the Ideal Gas Equation, to get a volume equivalent to this molar quantity:

V=(nRT)/P=(2.95*molxx0.0821*L*atm*K^-1*mol^-1xx307*K)/((781*mm*Hg)/(760*mm*Hg*atm^-1)

$\cong 72 \cdot L$

What mass of $\text{dioxygen}$ does this quantity constitute?

Again, I stress the unreality of measurements of $781 \cdot m m \cdot H g$ with respect to pressure.