Question

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?

Answer 1

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*L`.

Explanation:

We need a stoichiometrically balanced equation:

`2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)`

`"Moles of aluminum metal"` `=` `(53.1*g)/(27.0*g*mol^-1)=1.97*mol`

Thus, we need `1.97*molxx3/2*mol` `"dioxygen gas"` `=` `2.95*mol`.

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)`

`~=72*L`

What mass of `"dioxygen"` does this quantity constitute?

Again, I stress the unreality of measurements of `781*mm*Hg` with respect to pressure.