Question

A mixture of methane gas, CH4(g) , and pentane gas, C5H12(g) , has a pressure of 0.5239 atm when placed in a sealed container. The complete combustion of the mixture to carbon dioxide gas, CO2(g) , and water vapor, H2O(g) , was achieved by adding exactly enough oxygen gas, O2(g) , to the container. The pressure of the product mixture in the sealed container is 2.429 atm . Calculate the mole fraction of methane in the initial mixture, assuming the temperature and volume remain constant.

Answer 1

Answer:0.86

Explanation:

A mixture of methane gas, CH4(g), and pentane gas, C5H12(g), has a pressure of 0.5922 atm when placed in a sealed container.Lets suppose initially partial pressure of CH4 is x and C5H12 is y

x+y=0.5922-----i

1.CH4[g] + 2O2[g] -> CO2[g] + 2 H2O[g]

2.C5H12 + 8O2 → 5CO2 + 6H2O

partial pressure of CH4 is x produce xatm of CO2 and 2xatm of H2O

x + 2x = 3 atm

partial pressure of C5H12 is y produce 5yatm of CO2 and 6yatm of H2O

5y+6y=11y

so the final partial pressure of product mixture will be

3x + 11y = 2.429atm ------ii

solve eq i for x and put in ii

x=0.5922-y -----iii

3(0.5922-y)+ 11y = 2.429atm

y= 0.8155atm

put in iii

x=0.5922-0.8155

x= 0.51atm (Partial pressure of CH4)

mol fraction of CH4=Partial pressure/

Total pressure

=0.51/0.5922

=0.86.