ΔH∘ for, 2Al+Fe2O3⟶2Fe+Al2O3 is _____.Given that standard heat enthalpy of Fe2O3 and Al2O3 are -196.5 and -399.1 kcal.

ΔHReaction ∘=ΔHProducts ∘−ΔHReatants ∘=2×ΔHFe∘+ΔHAl2O3∘−2×ΔHAl∘+ΔHFe2O3∘∵ΔH∘ of free elements is zero, i.e., ΔHFe∘=0;ΔHAl∘=0=2×0+(−399.1)−[2×0+(−196.5)]∴ΔH∘=−202.6kcal