A mixture of helium gas and argon gas occupies 80.0 L at 398 K and 3.50 atm. If the mass of helium gas is equal to the mass of argon gas in the mixture, how many moles of helium does the mixture contain

7.79 moles

Explanation:

Let the mass of helium gas = Mass of argon gas = x g

Moles of helium = moles

Moles of argon = moles

Total moles =

Given that:

Temperature = 398 K

V = 80.0 L

Pressure = 3.50 atm

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

x=31.16 g

Moles of helium = 31.16 / 4 = 7.79 moles