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

  • Thread starter Thread starter pivoxa15
  • Start date Start date
  • Tags Tags
    Per
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
4 replies · 3K views
pivoxa15
Messages
2,250
Reaction score
1

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
 
Physics news on Phys.org

Discussion

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 [itex]\Delta H_c[/itex] (enthalpy of combustion), then it is per mole of the hydrocarbon being combusted. If it is specified as [itex]\Delta H_r[/itex] (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