# Per mol of what?

## 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?

## 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

## 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?

## 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
Gold Member

## 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?

## 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.

Gokul43201
Staff Emeritus