Mixture Composition Calculation for Combustion of Methane and Propane

[sitam]

Say we have a mixture of 10 mL of methane and propane. For a complete combustion, we need 41 mL of O2. Calculate the composition of the mixture in terms of volume, if atmospheric temperature and pressure are constants.

The equations are:
CH4 + O2 --> CO2 + H2O
C3H8 + O2 --> CO2 + H2O

I do not understand the question: am I asked to calculate a percentage of volumes, e.g. VCH4/VC3H8, or simply to find the volumes of CH4 and C3H8 separately (if so, how?)? Am I considering the reactives or the products of the reaction?

These calculations must come in somewhere in the solution:
I know that all gazes have the same molecular volume at normal temperature and pressure: 22.4L/mol. Therefore I can determine the quantity of moles of each component. In the initial mixture (before reaction), there is:
0.224mol of CH4 + C3H8
0.918mol of O2.

Thx. for helping!

 

[Bystander]

composition

volumes of CH4 and C3H8

You want the volume fraction of the initial mixture, (methane/(methane + propane)), or (propane/(methane + propane)).

0.224mol of CH4 + C3H8

This is correct, given the ideal gas constraint.

41 mL of O2

CH4 + O2 --> CO2 + H2O
C3H8 + O2 --> CO2 + H2O

Balance these two equations.
Solve the system of two equations in two unknowns, number of moles ∝ 10 and number of moles ∝ f(41).