A mixture of argon and xenon gases is maintained in a 8.89 L flask

Question

A mixture of argon and xenon gases is maintained in a 8.89 L flask at a pressure of 2.36 atm and a temperature of 22 °C. If the gas mixture contains 16.2 grams of argon, the number of grams of xenon in the mixture is _______g.

Answer 1

The total pressure (P) can be calculated using ideal gas equation: PV = nRT,

 

for the above problem n is the total number of moles, thus

 

n =PV/RT

 

P = 2.36 atm, V = 8.89 L, T = 22^oC + 273 = 295 K, R = 0.0821 L.atm/mol.K (it is a constant).

 

I will let you to do the calculation.  you should get n =  0.866 mol

 

Thus,  0.866 = mol of Ar + mol of Xe

 

There is 16.2 g of Ar =  0.406 mol (use the molar mass of Ar and convert 16.2 g into mol)

 

0.866 mol = 0.406 mol + mool Xe

 

mol Xe = 0.866 mol - 0.406 mol = 0.46 mol

 

Thus, now using the molar mass of Xe convert 0.46 mol to gram.

 

FYI: the answer is 60.4 g