Answer:
[tex]\Delta H_{rxn}=-1234.782kJ[/tex]
Explanation:
[tex]\Delta H_{rxn}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})][/tex]
Where [tex]n_{i}[/tex] and [tex]n_{j}[/tex] are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).
[tex]\Delta H_{f}^{0}[/tex] is standard heat of formation.
So, [tex]\Delta H_{rxn}=[2mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[3mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{2}H_{5}OH)_{l}]-[3mol\times \Delta H_{f}^{0}(O_{2})_{g}][/tex]
or, [tex]\Delta H_{rxn}=[2mol\times -393.509kJ/mol]+[3mol\times -241.818kJ/mol]-[1mol\times -277.69kJ/mol]-[3mol\times 0kJ/mol][/tex]
or, [tex]\Delta H_{rxn}=-1234.782kJ[/tex]