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# What is the enthalpy change ($\Delta$H) in kJ for the following reaction?2Al(s) + Fe$_2$O$_3$(s) $\longrightarrow$ 2Fe(s) + Al$_2$O$_3$(s)Use the enthalpy changes for the combustion of aluminum and iron:2Al(s) + $\frac{3}{2}$O$_2$(g) $\longrightarrow$ Al$_2$O$_3$(s) $\,\,\,\,\,\,\,\,\,\,\,\,\Delta$H = -1676.0 kJ2Fe(s) + $\frac{3}{2}$O$_2$(g) $\longrightarrow$ Fe$_2$O$_3$(s) $\,\,\,\,\,\,\,\,\,\,\,\,\Delta$H = -822.1 kJ

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To calculate the enthalpy change of the final equation, the given equations must be summed. In the given state by the sum of the equations it is impossible to obtain the final one. Therefore, the second equation will be reversed, thus reversing the sign of the enthalpy change of that equation. The sum of the equations subtracts the molecules that appear on different sides of the final equation. the default enthalpy changes are simply summed. This principle is called Hess's law.

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