Many cigarette lighters contain liquid butane, C4H10(l). Using standard enthalpies of formation, calculate the quantity of heat produced when 1.2 g of butane is completely combusted in air under standard conditions. Answer in kJ.
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1 mol of C4H10 needs to produce 4 mols of CO2 and 5 mols of H2O. If you now count up how many mols of O are required then this comes to 8+5=13. Oxygen molecules have 2 O atoms so to balance this equation you get: 2C4H10+13O2 --> 8CO2 + 10H2O
Now you need to apply the Enthalpies of each of the four different molecules. Everything on the left is heat consumption or negative heat release and everything on the right is heat release.
So now get your tables out that define the enthalpies of form or refer to the link that I have given. C4H10 = 30.1 Kj/mol CO2 = 393.5 Kj/mol H2O = 285.5 Kj/mol O2 is neutral.
Now work out how many mols you have. Molecular weight of butane is 58 so 1.2/58 gives 0.0207 mols.
Now you can do the calculation to work out heat release. On the left (pre-combustion) you have 0.0207x30.1=0.6228 Kj On the right you have 4x0.0207x393.5 + 5x0.0207x285.5 = 62.13Kj
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