How Much Butane is Needed for 33440J of Energy? Combustion Equation and Solution

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Discussion Overview

The discussion revolves around the calculation of the mass of butane required to produce a specific amount of energy (33440J) through combustion. Participants explore the interpretation of the enthalpy change (\( \Delta H \)) associated with the combustion reaction and its implications for stoichiometry and energy calculations.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants question the interpretation of \( \Delta H = -5760 \text{ kJ/mol} \) and whether it refers to the complete reaction of butane or just one mole of butane.
  • There is a discussion about whether the enthalpy change is specified per mole of butane or per the entire reaction, with some suggesting it should be per mole of the hydrocarbon combusted.
  • One participant notes a discrepancy in the interpretation of energy release in a different combustion reaction involving ethane, suggesting that the conventions used in the textbook may not align with their calculations.
  • Another participant expresses confusion over the application of stoichiometric coefficients in the combustion equations and how they relate to the energy produced.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the enthalpy change and its application to the combustion of butane. There is no consensus on the correct approach to calculating the mass of butane needed for the specified energy output.

Contextual Notes

Participants highlight potential ambiguities in the definitions and units used for enthalpy changes, which may affect their calculations. The discussion reflects uncertainty regarding the application of stoichiometric principles in energy calculations.

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Homework Statement


The combustion of the fuel gas butane is given by:

2(C4H10) +13(O2) --> 8(CO2) + 10(H2O) del(H)= –5760kJ/mol

What mass of butane would be needed to produce 33440J of energy?



Homework Equations





The Attempt at a Solution


First of all how do you interpret del(H)= –5760kJ/mol
Per mol of what?

I assume a mol is the complete reaction of
[2(C4H10) +13(O2) --> 8(CO2) + 10(H2O)]

33440J/(5760kJ/mol)=0.0058mol of the reaction

So a total of 0.0058*2=0.0116 moles of butane?

Which would be 0.012*(12.01*4 + 10.1)=0.675g of butane but the answers suggested 0.337g which is a half of what I worked out. Hence they didn’t multiply by 2. This confusion comes back to my first question. ‘Per mol of what?’ In del(H)= –5760kJ/mol
 
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pivoxa15 said:

Homework Statement


The combustion of the fuel gas butane is given by:

2(C4H10) +13(O2) --> 8(CO2) + 10(H2O) del(H)= –5760kJ/mol

What mass of butane would be needed to produce 33440J of energy?

Homework Equations


The Attempt at a Solution


First of all how do you interpret del(H)= –5760kJ/mol
Per mol of what?

No. of moles is the stoichiometric coeff. in a balanced eqn.

oh BTW, how is the eqn represented in the text...i mean like
2(C4H10) +13(O2) --> 8(CO2) + 10(H2O) + 5760 J
or
is the energy evolved given separately like 5760 J/mol
In case 1, 2 mol give 5760 J while in case 2, 1 mol gives 5760 J
 
Last edited:
pivoxa15 said:

Homework Statement


The combustion of the fuel gas butane is given by:

2(C4H10) +13(O2) --> 8(CO2) + 10(H2O) del(H)= –5760kJ/mol

What mass of butane would be needed to produce 33440J of energy?



Homework Equations





The Attempt at a Solution


First of all how do you interpret del(H)= –5760kJ/mol
Per mol of what?
If it is specified as \Delta H_c (enthalpy of combustion), then it is per mole of the hydrocarbon being combusted. If it is specified as \Delta H_r (enthalpy of reaction), then it would not come in units of kJ/mol, and would simply be in units of kJ. This would indicate the question is talking about an enthalpy of combustion, so it's for 1 mole of butane.
 
That's strange, in another part of the book, it stated the equation
2(C2H6) + 7(O2) --> 4(CO2) + 6(H2O) del(H)=-2856kJ/mol

and explained, "This equation means that if 2 moles of C2H6 reacts with 7 moles of O2, then 2856 kJ of heat energy will be released." But the answers worked out didn't go by this convention.
 
pivoxa15 said:
That's strange, in another part of the book, it stated the equation
2(C2H6) + 7(O2) --> 4(CO2) + 6(H2O) del(H)=-2856kJ/mol

and explained, "This equation means that if 2 moles of C2H6 reacts with 7 moles of O2, then 2856 kJ of heat energy will be released."
This is definitely WRONG. Which text is this?

But the answers worked out didn't go by this convention.
Did the given answers use the correct definition?
http://www.answers.com/topic/standard-enthalpy-change-of-combustion
 

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