Mole fraction of methane and ethane?

Question

a mixture of methane CH4 and ethane C2H6 is stored in a container at 294 mmHg. The gases are burned in air to form CO2 and H2O. If the pressure of CO2 is 343 mmHg measured at the same temp and volume as the origional mixture, calculate the mole fraction of the gases. a) mole fraction of methane?  b) mole fraction of ethane?

Answer 1

The first step is to write the two balanced combustion equations:

 

      CH₄ + 2 O₂ ==>    CO₂ + 2 H₂O      C₂H₆ + (7/2) O₂  ==>   2 CO₂ + 3 H₂O

 

    Let the number of moles of methane be a  and the number of moles of ethane be b.

    Then before combustion, there are a + b moles of gas.    This amount is then mixed with (excess) oxygen and burned to produce   a + 2 b moles of CO₂ .  

 

   According to the ideal gas equation    294 = (a+b) R T / V       (situation before O₂ is added and combusted)

 

   Also according to the ideal gas equation  343 = (a + 2b) RT /V   (situation after 343 is partial pressure of CO₂

 

   Dividing the second equation by the first results in 1.1666 = (a + 2b) /(a +b).   This equation can be solved

in algebra to give a/b = 11.        It then follows that b/(a+b) =  1/12 and  a/(a+b) =  11/12.

 

These are the desired mole fractions of ethane and methane.