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What is the standard enthalpy change for the combustion of 3 moles of propane gas (C3H8) at standard conditions? Express your answer in kJ/mol of propane burned.

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To calculate the standard enthalpy change for the combustion of propane, we first need to write the balanced chemical equation for the combustion reaction:C3H8 g  + 5O2 g   3CO2 g  + 4H2O l Next, we need to find the standard enthalpies of formation  Hf  for each compound involved in the reaction. The values are as follows:Hf [C3H8 g ] = -103.85 kJ/molHf [O2 g ] = 0 kJ/mol  since it's an element in its standard state Hf [CO2 g ] = -393.5 kJ/molHf [H2O l ] = -285.8 kJ/molNow, we can calculate the standard enthalpy change for the reaction  H  using the following equation:H = [sum of  moles  Hf of products ] - [sum of  moles  Hf of reactants ]H = [ 3  -393.5  +  4  -285.8 ] - [ 1  -103.85  +  5  0 ]H =  -1180.5 + -1143.2  -  -103.85 H = -2323.7 + 103.85H = -2219.85 kJSince this is the enthalpy change for 3 moles of propane, we need to divide by 3 to get the enthalpy change per mole of propane:H  per mole of propane  = -2219.85 kJ / 3H  per mole of propane  = -739.95 kJ/molSo, the standard enthalpy change for the combustion of 1 mole of propane gas  C3H8  at standard conditions is approximately -739.95 kJ/mol.

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