Question

A mixture containing nitrogen and hydrogen weighs 3.48 g and occupies a volume of 7.47 L at 296 K and 1.02 atm. Calculate the mass percent of these two gases. Assume ideal-gas behavior.

Answer 1

there is 2% of hydrogen and 98% of nitrogen (mass percent)

Step-by-step explanation:

assuming ideal gas behaviour

P*V=n*R*T

n= P*V/(R*T)

where P= pressure=1.02 atm , V=volume=7.47 L , T=absolute temperature= 296 K and R= ideal gas constant = 0.082 atm*L/(mole*K)

thus

n= P*V/(R*T) = 1.02 atm*7.47 L/( 296 K * 0.082 atm*L/(mole*K)) = 0.314 moles

since the number of moles is related with the mass m through the molecular weight M

n=m/M

thus denoting 1 as hydrogen and 2 as nitrogen

m₁+m₂ = mt (total mass)

m₁/M₁+m₂/M₂ = n

dividing one equation by the other and denoting mass fraction w₁= m₁/mt , w₂= m₂/mt , w₂= 1- w₁

w₁/M₁+w₂/M₂ = n/mt

w₁/M₁+(1-w₁) /M₂ = n/mt

w₁*(1/M₁- 1/M₂) + 1/M₂ = n/mt

w₁= (n/mt- 1/M₂) /(1/M₁- 1/M₂)

replacing values

w₁= (n/mt- 1/M₂) /(1/M₁- 1/M₂) = (0.314 moles/3.48 g - 1/(14 g/mole)) /(1/(1 g/mole)-1/(14 g/mole))= 0.02 (%)

and w₂= 1-w₁= 0.98 (98%)

thus there is 2% of hydrogen and 98% of nitrogen