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Calculate the heat released when 1 mole of ethane (C2H6) is combusted in oxygen gas to form carbon dioxide and water, given the following enthalpy of combustion values: ΔH°comb(C2H6) = -1560 kJ/mol ΔH°comb(O2) = -393.5 kJ/mol ΔH°f(CO2) = -393.5 kJ/mol ΔH°f(H2O) = -285.8 kJ/mol

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To calculate the heat released during the combustion of 1 mole of ethane, we can use the balanced chemical equation for the combustion reaction and the given enthalpy values.The balanced chemical equation for the combustion of ethane is:C2H6 + 7/2 O2  2 CO2 + 3 H2OUsing the given enthalpy values, we can calculate the heat released  H  for the reaction using Hess's Law:H =  Hf products  -  Hf reactants For the products, we have 2 moles of CO2 and 3 moles of H2O:Hf products  = 2 * Hf CO2  + 3 * Hf H2O Hf products  = 2 *  -393.5 kJ/mol  + 3 *  -285.8 kJ/mol Hf products  = -787 kJ +  -857.4 kJ Hf products  = -1644.4 kJFor the reactants, we have 1 mole of C2H6 and 7/2 moles of O2. Since the enthalpy of formation for an element in its standard state  O2  is zero, we only need to consider the enthalpy of combustion for ethane:Hf reactants  = Hcomb C2H6 Hf reactants  = -1560 kJNow, we can calculate the heat released  H  for the reaction:H = Hf products  - Hf reactants H = -1644.4 kJ -  -1560 kJ H = -84.4 kJSo, the heat released when 1 mole of ethane is combusted in oxygen gas to form carbon dioxide and water is -84.4 kJ.

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