An oxygen gas container has a volume of 20.0 L. How many grams of oxygen are in the container if the gas has a pressure of 876 mmHg at 23 C?

Question

Owen Ambrose · Accepted Answer

There are #30.4# grams of oxygen in the sample.

Use the ideal gas law. We are given the pressure, the temperature and the volume. The gas constant will be #62.36# #PV = nRT# We're looking for #n#, the number of moles in the sample. Note that #23˚C= 23 + 273 = 296K# #(20.0 L)(876 mmHg) = n(62.36 mmHg L k^-1 mol^-1)296K# If we cancel all units we're left with #n = 0.949 # mol Knowing that #"molar mass" = ("number of grams per sample")/("number of moles")# And that the molar mass of oxygen is #32.00 g/(mol)#, we can solve for number of grams in sample. #32.00 g/(mol) * 0.949 mol = g# #g = 30.4 " grams"# Notice I rounded to #3# sig figs. Hopefully this helps!

Owen Ambrose · Answer

undefined To find the mass of oxygen in the container, use the ideal gas law equation:

$PV = nRT$

Rearrange the equation to solve for moles (n):

$n = \frac{{PV}}{{RT}}$

$P = 876 \, \text{mmHg} = 876 \times 1.000 \, \text{kPa/mmHg} = 115.808 \, \text{kPa}$

$V = 20.0 \, \text{L}$

$T = 23 + 273.15 = 296.15 \, \text{K}$

R = 8.314 J/(mol*K)

Substitute the values into the equation and solve for moles:

$n = \frac{{115.808 \, \text{kPa} \times 20.0 \, \text{L}}}{{8.314 \, \text{J/(mol*K)} \times 296.15 \, \text{K}}}$

Calculate moles:

$n = 0.938 \, \text{mol}$

Use the molar mass of oxygen to convert moles to grams:

Molar mass of oxygen (O₂) = 2(16.00 g/mol) = 32.00 g/mol

$m = n \times \text{molar mass}$

$m = 0.938 \, \text{mol} \times 32.00 \, \text{g/mol} = 30.016 \, \text{g}$

Therefore, there are approximately 30.02 grams of oxygen in the container.