Complete Combustion - which hydrocarbon will consume more O2

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Discussion Overview

The discussion revolves around determining which hydrocarbon (propane, ethane, methane, or pentane) will consume the greatest mass of oxygen during complete combustion. Participants are exploring the stoichiometry of combustion reactions and the implications of molar mass on oxygen consumption.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that a greater number of moles of hydrocarbon would require more oxygen for combustion, leading to the assumption that methane, having a lower molar mass, would need more oxygen.
  • Another participant provides balanced chemical equations for the complete combustion of each hydrocarbon and calculates the moles of O2 required for each, questioning the correctness of their calculations.
  • Calculated values for moles of O2 needed for combustion are presented, with one participant noting that the hydrocarbon in answer D (pentane) does not appear to require the maximum amount of O2 based on their calculations.
  • A later reply concurs with the calculations and suggests that methane (answer C) should be the correct answer, implying a potential error in the provided solution.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correct answer to the problem, with some supporting the original solution (D) and others arguing for methane (C) based on their calculations. The discussion remains unresolved as participants continue to question the accuracy of their findings and the provided solution.

Contextual Notes

Participants have not reached a consensus on the correct answer, and there are indications of possible algebraic errors in the calculations. The discussion reflects uncertainty regarding the stoichiometric relationships in the combustion reactions.

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Homework Statement


one gram of each of the following gases is introduced into a 10 L container at 25 degrees C

a) propane
b) ethane
c) methane
d) pentane

which gas will consume the greatest mass of oxygen upon complete combustion?

The solutions says that the right answer is D. I do not know why

Homework Equations


no equations. just chemical ones.

The Attempt at a Solution



My logic is that if I have greater number of moles of hydrocarbon, I will need more oxygen to consume.

I thus thought that since methane has a lesser molar mass, we have greater moles, so we will need greater moles [which leads to a higher mass] of O2 gas needed.

Could anyone please advise on how to solve this?

Thanks!
 
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Start with writing out balanced chemical equations for the complete combustion of each of these hydrocarbons.
 
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OK.

My equations are as following
  • propane: C3H8 + 5O2 --> 3CO2 + 4H2O
  • ethane: 2C2H6 + 7O2 ----> 4CO2 + 6H2O
  • CH4 + 2O2 ---> Co2 + 2H2O
  • C5H12 + 8O2 ----> 5CO2 + 6H2O
I believe these equations are correct. right?

I just did some stoichiometry with approximate values, and I got 0.111 moles of O2 needed to combust the hydrocarbon in answer D.

My other calculated numbers are:
a) 0.1136 mol O2 needed
b) 0.1167 mol o2 needed
c) 0.125 mol o2 needed
d) 0.111 mol o2 needed.

So far, it seems that the hydrocarbon in answer d does not have the maximum need for O2 moles.

Could I have made an algebraic error?

For example, a sample calculation for my number for part A is:

(1 g C3H8)/(44 g C3H8) * ( 5 mol O2 / 1 mol C3H8) = 0.1136.

I could be wrong.

Please advise and thanks for your help!
 
I did the calculations as well and got the same numbers as you. The answer should be C (methane), and the book is probably wrong.
 
Wow. This seems really strange.

Thanks so much for the help!
 

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