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?

1 Answer
Dec 15, 2016

Answer:

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.