Calculate the enthalpy change for the thermite reaction: 2al(s)+fe2o3(s)→2fe(s)+al2o3(s), δh∘rxn=−850 kj when 10.0 mol of al undergoes the reaction with a stoichiometrically equivalent amount of fe2o3.

From the balanced equation for this reaction: 
2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s)
so from this balanced equation, we can know that:
2 moles of Al react with 1 mole of FeO3 to give 850 Kj 
So the energy is given by 10 mol of Al should be calculated from
2 mol Al → -850 KJ
10 mol Al→ ???
and it is obvious that as the number of moles increases so the energy will be higher. 
∴ ΔH°rxn= -850 kj * 10 mol of Al / 2 mol of Al
            = -4250 KJ

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