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?

Question

Luna Chen · Accepted 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#.

A stoichiometrically balanced equation is required: #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#. Now, all we have to do is apply the Ideal Gas Equation to obtain the volume equal 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 is the mass of "dioxygen" that this quantity is made up of? Again, I stress the unreality of measurements of #781*mm*Hg# with respect to pressure.

Noah Aldrich · Answer

undefined To find the volume of O₂ gas, use the ideal gas law:

$$ V = \frac{nRT}{P} $$

Where:
$ V $ = volume
$ n $ = moles of gas
$ R $ = ideal gas constant (0.0821 L·atm/mol·K)
$ T $ = temperature in Kelvin (273 + degrees Celsius)
$ P $ = pressure

1. Find moles of Al using its molar mass.
2. Write and balance the chemical equation for the reaction.
3. Use stoichiometry to find moles of O₂.
4. Apply the ideal gas law to find the volume of O₂.