This problem is providing us with the enthalpy of formation of Fe2O3 (s) and the enthalpy of reaction when it reacts with Al to form Al2O3 (s) and Fe. Thus, the enthalpy of formation of Al2O3 (s) is required and found to be -1675.7 kJ/mol as follows.
In chemistry, the heat involved in a chemical process can be quantified via the difference in the overall enthalpy of products minus that of reactants:
[tex]\Delta H_r=\Delta H_{products}-\Delta H_{reactants}[/tex]
Now, with the given reaction:
[tex]Fe_2O_3 (s) + 2 Al (s) \rightarrow 2 Fe (s) + Al_2O_3 (s)[/tex]
One can see both Al and Fe have zero as their enthalpies of formation for they are elements. In such a way, one can rearrange the previous formula as:
[tex]\Delta H_r=\Delta H_{Al_2O_3}-\Delta H_{Fe_2O_3}[/tex]
And hence plug in the given energies in order to find the required:
[tex]-851.5kJ/mol=\Delta H_{Al_2O_3}-(-824.2kJ/mol)\\\\\Delta H_{Al_2O_3}=-851.5kJ/mol+(-824.2kJ/mol)\\\\\Delta H_{Al_2O_3}=-1675.7kJ/mol[/tex]
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