What Are the Products of Acetylene Combustion in Air?

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Discussion Overview

The discussion revolves around the combustion of acetylene (C2H2) in air, specifically focusing on the products formed during this process. Participants explore the basic chemical equation for the combustion and the role of nitrogen in the reaction, considering both complete combustion and the implications of the fuel-to-air ratio.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant seeks clarification on the basic chemical equation for the combustion of acetylene, questioning whether the products are carbon dioxide (CO2), water (H2O), and nitrogen (N2).
  • Another participant asserts that nitrogen should not be included in the equation as a product, suggesting it acts as a spectator.
  • A different participant counters that nitrogen is not a spectator and emphasizes that the products can vary significantly based on the fuel-to-air ratio, mentioning that various nitrogen oxides (NO(x)), nitrous oxide (N2O), and cyanide (CN-) can be produced in a stoichiometric flame.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the role of nitrogen in the combustion products, with some asserting it is a spectator while others argue it plays a significant role in the reaction outcomes.

Contextual Notes

The discussion highlights the complexity of combustion reactions and the dependence of products on specific conditions, such as the fuel-to-air ratio, which remains unresolved.

Jacob87411
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Acetylene (C2H2) is burned with a stoichiometric amount of air during a combustion procss. Assuming complete combustion, determine the air-fiel ratio on a mass and mole basis.

I'm not asking for help on the actual question I just don't know what the basic chemical equation is?

C2H2 + (O2+3.76N2) -> what?

Does it yield CO2, H20 and N2? Thanks for any help
 
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Jacob87411 said:
Acetylene (C2H2) is burned with a stoichiometric amount of air during a combustion procss. Assuming complete combustion, determine the air-fiel ratio on a mass and mole basis.

I'm not asking for help on the actual question I just don't know what the basic chemical equation is?

C2H2 + (O2+3.76N2) -> what?

Does it yield CO2, H20 and N2? Thanks for any help

You must start with these facts: Your reactants are acetylene and oxygen. Your products are carbon dioxide and water. Nitrogen is probably not going to be involved, but I'm not absolutely certain about this. Try balancing the equation from the point of view of the reactants and products just stated.
 
Well my question wasn't with balancing it, I just wanted to confirm what the product were, not how much of each. So CO2, H20 and N2 are the products?
 
Treat N2 as a spectator - don't put it into the equation.

Borek
 
Last edited by a moderator:
I know this a bit of a dead thread, however:
Nitrogen most certainly is not a spectator. The products depend completely on the F:A ratio. Many NO(x)'s, N2O, and CN- are some of the products running a "stoichiometric" flame. All produce molecular bands in FES spectra.
 

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