Question

In a right pressure vessel, a mixture was prepared with only CH4 (g) and O2 (g). The mole fraction of the CH4 g was initially 0.50. A spark then initiated combustion to CO2 (g) and H2O (g). What is the mole fraction of CO2 (g) in the pressured vessel after combustion? Assume H2O remains in the gaseous phase.

Answer 1

Ans :

The combustion reaction is given as :

CH4 + 2O2 = CO2 + 2H2O

Let the total amount of moles in the reaction mixture be 1 mol

Mole fraction of CH4 = 0.50

So number of mol of methane in mixture = 0.50 mole

and number of mol of O2 = 0.50 mole

0.50 mol O2 will burn : 0.50 mol /2 = 0.25 mol CH4

So number of moles of CO2 and H2O produced will be 0.25 mol and 0.5 mol respectively.

mol CH4 left = 0.50 mol - 0.25 mol = 0.25 mol

The in the mixture we have : 0.25 mol CH4, 0.25 mol CO2 and 0.5 mol H2O

total moles = 0.25 mol + 0.25 mol + 0.5 mol = 1.0 mol

Mole fraction of CO2 = mole CO2 / total mol

= 0.25 mol / 1.0 mol

= 0.25