Many cigarette lighters contain butane, C4H10(l). Using standard enthalpies of f

Question

Many cigarette lighters contain butane, C4H10(l). Using standard enthalpies of formation, calculate the quantity of heat produced when 5.00g of butane is completely combusted in air under standard conditions.

Explanation / Answer

2C4H10 + 5O2 --> 8CO2 + 10H2O this requires an Hformation table Hf for O2 = 0 since it is a pure element CO2 = -393.5kJ/mole H2O = -285.8kJ/mole C4H10 = -124.7kJ/mole 5gbutane / 58g/mole = 0.088moles, this yields 0.313moles CO2 and 0.392moles H2O deltaHf = (124.7 x 0.078) + (-393.5 x 0.313) + (-285.8 x 0.392) = -225.47kJ

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