A mixture of neon and argon gases is maintained in a 7.81 L flask at a pressure of 2.97 atm

Question

A mixture of neon and argon gases is maintained in a 7.81 L flask at a pressure of 2.97 atm and a temperature of 68 °C. If the gas mixture contains 9.75 grams of neon, the number of grams of argon in the mixture is ???  g.

Answer 1

From the ideal gas law PV = nRT we can find the total moles of gas (n). After finding that, we can subtract the moles of neon to find the moles of argon and then convert moles of argon to grams of argon.

n = PV/RT

n = moles = ?

P = pressure = 2.97 atm

V = volume = 7.81 L

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K =68ºC + 2713 = 341K

n = (2.97)(7.81)/(0.0821)(341) = 0.829 moles of gas

moles Ne = 9.75 g x 1 mol/20.2 g = 0.483 moles Ne

moles Ar = 0.829 - 0.483 = 0.346 moles Ar

mass Ar = 0.346 mol x 39.9 g/mol = 13.8 g Ar in the mixture

Answer 2