To solve, use Appendix II in your textbook (containing standard heats of formation)
ΔH°- ?
ΔHf(Fe) = 0
ΔH (Al) = 0
ΔHf (Fe2O3) = -822.2 kJ/mol
ΔHf (Al2O3) = -1676 kJ/mol
ΔH° = n ΔHf (prod.) – m ΔHf (reag.)
ΔH° = -1676 – (-822.2) = -853.8 kJ/mol
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