October 20, 2021, 01:23:39 PM
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### Topic: Stoichiometric ratio  (Read 436 times)

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#### Pez

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• Mole Snacks: +0/-0 ##### Stoichiometric ratio
« on: June 07, 2021, 04:13:04 PM »
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...)

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?

Perry

« Last Edit: June 07, 2021, 04:29:15 PM by Pez »

#### Orcio_Dojek ##### Re: Stoichiometric ratio
« Reply #1 on: June 07, 2021, 04:46:52 PM »
Quote
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)).

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

You need to assume some mass (for example 1000 g) and then convert mass into volume.

Then write all equations (burning of CH4, C3H8 and H2S - which gives SO2 not elemental sulfur).

Knowing how many liters of air is needed for one liter of component, rest is very simple....

#### Borek ##### Re: Stoichiometric ratio
« Reply #2 on: June 08, 2021, 03:09:09 AM »
Note: while assuming some gas mass is a useful trick, yielding a correct answer, it is actually not necessary. If you start with mass m and do calculations on symbols, in the end m will cancel out leaving just the answer.
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