Hi All,

Perhaps slightly different to the normal questions here as it's from an Engineering background.

We've been given the following formula based on a combustion test with a natural gas with Methane as its main content.

CH4 + 2(O2 + 3.76N2) = CO2 + 2H2O + 2 x 3.76N2

And therefor one part of methane will require 9.52 for AFR stoichiometric combustion. (I think I follow this so far...)

The question we've been asked to answer is:

**Calculating the stoichiometric air/fuel ratio of natural gas (assuming it has 85% methane, 6%**

propane, 5% carbon dioxide, 3% nitrogen, 1% hydrogen sulphide (all by mass)).

I believe I have balanced the equation to produce:

CH4+C3H8+CO2+N2+H2S +7.5(O2+3.76N2) -> 5CO2+7H20+28.2N2+S

So my question (I think) is.... How do I work out the AFR stroichiometric combustion of this combined fuel?

Thanks in advance!

Perry