Question

A mixture of methane gas, CH4(g), and pentane gas, C5H12(g), has a pressure of 0.5015 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 2.421 atm. Calculate the mole fraction of methane in the initial mixture assuming the temperature and volume remain constant. Xch4=

Answer 1

CH4 gives 1CO2 and 2H2O , total 3 gas molecules per 1CH4

C5H12 gives 5CO2 and 6H2O total 11 gas molecules per 1C5H12

Initially P = 0.5015 atm , we use PV = nRT equation

0.5015 x V = ( n1+ n2) x RT ................(1) where CH4 moles = n1 , C5H12 moles = n2

after combustion P = 2.421 atm , hence

2.421 x V = ( 3n1+ 11n2) x RT .....(2)

now ( 2/1)gives 2.421/0.5015 = ( 3n1+ 11n2) / ( n1+n2)

4.8275n1 + 4.8275 n2 = 3n1+11n2

1.8275 n1 = 6.1725 n2     hence n2 = 0.296 n1

mol fraction of CH4 = n1 / ( n1+n2) = n1 / ( n1+0.296n1) = n1 / ( 1.296) n1 = 0.7716