Fe2O3(s) + 2Al(s) → Al2O\(_3...
Fe2O3(s) + 2Al(s) → Al2O3 + 2Fe(s). If the heats of formation of Al2O3 and Fe2O3 are - 1670 kJmol−1 and - 822 kJmol−1 respectively, the enthalpy change in kJ for the reaction is ?
+2492
+848
-848
-2492
Correct answer is C
Heat of formation = Heat of products - Heat of reactants
= - 1670 - (- 822)
= - 1670 + 822
= - 848 kJ