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Enthalpy Problem. Please Help?
Many cigarette lighters contain liquid butane, C4H10(l). Using standard enthalpies of formation, calculate the quantity of heat produced when 1.7 g of butane is completely combusted in air under standard conditions.
Can someone show me how to set it up?
Thanks
1 Answer
- ChemTeamLv 710 years agoFavorite Answer
This is the equation you need:
C4H10 + (9/2)O2 ---> 2CO2 + 5H2O
You need to look up the enthalpy of formation for C4H10, CO2 and H2O. The enthalpy of formation for O2 is zero.
Then you need to calculate the enthalpy of combustion for one mole of C4H10 using Hess' Law. Here are some examples:
http://www.chemteam.info/Thermochem/HessLawIntro2....
Look at examples 1 and 2.
Once you have the enthalpy of combustion (and it will be in kJ/mol), you need to do one more thing.
(enthalpy of combustion) times (1.7 g / molar mass of C4H10) = the answer to your question
This term:
(1.7 g / molar mass of C4H10)
calculates the moles of C4H10 in 1.7 g of C4H10
