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Per mol of what?

  • Thread starter pivoxa15
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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?



2. Homework Equations



3. 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
 

Answers and Replies

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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?



2. Homework Equations



3. 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:
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
6,987
14
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?



2. Homework Equations



3. 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.
 
2,234
1
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.
 
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
6,987
14
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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