The partial pressure of CH4 is 0.175 atm

Question

A. If the partial pressure of oxygen is 0.250 atm, what is the mole fraction of each gas in the mixture?

 

B. If the mixture occupies a volume of 10.5L at 65°C how many moles of gas are in the mixture?

 

C. Calculate the grams of each gas in the mixture

Answer 1

(a)

Partial pressure CH4 = 0.175 atm

Partial pressure O2 = 0.250 atm

Total pressure = 0.175 atm + 0.250 atm = 0.425 atm

mole fraction CH4 = 0.175/0.425 = 0.412

mole fraction O2 = 0.250/0.425 = 0.588

 

(b)

PV = nRT

(0.425 atm)(10.5 L) = n(0.0821 Latm/Kmol)(338 K)

n = moles = 0.161 moles of gas present in mixture

 

(c)

moles CH4 = mole fraction x moles = 0.412 x 0.161 = 0.0663 moles 

mass CH4 = 0.0663 moles x 16 g/mole = 1.06 g CH4

 

moles O2 = mole fraction x moles = 0.588 x 0.161 = 0.0947 moles

mass O2 = 0.0947 moles x 32 g/mole = 3.03 g O2