Question

A volume of 18.0 L contains a mixture of 0.250 mole N2 , 0.250 mole O2 , and an unknown quantity of He. The temperature of the mixture is 0 ∘C , and the total pressure is 1.00 atm . How many grams of helium are present in the gas mixture?

Answer 1

Assume the gases follow the ideal gas law:
PV = nRT
P = Pressure = 1.00atm
V = Volume = 18.0L
n = # of moles = What you're trying to find
R = Ideal Gas Constant = .0821 (L * atm) / (mol * K)
T = Temperature (in Kelvin) = 273K

1.00atm * 18.0L = (# of moles) * .0821 (L * atm) / (mol * K) * 273K
(# of moles) = (1.00 * 18.0) / (.0821 * 273) = .803
You are already given that the mixture has .250 mole N2 and .250 mole O2. Therefore, the # of moles of Helium gas is .803 - .250 - .250 = .303

Now we figure out how many grams this is.
The molar mass of helium is 4.0026g/mol. Therefore, the mass of helium will be 0.303moles * 4.0026g/mol = 1.21g.

1.21 gm of helium presents in reaction mixture