What is the identity of the third gas

Question

A 50.00L sealed vessel is filled with a mixture of three gases at 25.0 degrees C. The mixture has a total mass of 137.860 g. and exerts a total pressure of 2.875 atm. if the mixture contains 4.00 moles of N2 and 1.00 mole of He, what is the identity of the third gas?

Answer 1

PV = nRT

P = pressure = 2.875 atm

V = 50.00 L

n = moles = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 25.0ºC + 273.15 = 298.15K

Solving for moles (n), we have...

n = PV/RT = (2.875)(50) / (0.0821)(298.15

n = 5.8726 moles

Since the N₂ and He occupy 4.00 + 1.00 moles = 5.00 moles, the third gas must be 0.8726 moles

In terms of mass, we have...

4.00 mols N₂ x 28.0135 g / mol = 112.054 g

1.00 mols He x 4.0026 g / mol = 4.0026 g

mass of third gas = 137.860 g - 112.054 g - 4.0026 g = 21.8034 g

21.8034 g / 0.8726 mols = 24.987 g/ mol = molar mass of unknown gas

I personally don't know what gas has a molar mass of 24.99 g/mol. I think my math is correct, but you should check it carefully. One would expect that answer to be a common gas but none of the common gases have this calculated molar mass. Good luck and let us know if you find the correct answer.