Calculate the enthalpy change for the thermite reaction: 2al(s)+fe2o3(s)→2fe(s)+al2o3(s), δh∘rxn=−850 kj when 6.00 mol of al undergoes the reaction with a stoichiometrically equivalent amount of fe2o3.

Answer is: the enthalpy change for the thermite reaction is -2550 kJ.

Balanced chemical reaction: Fe₂O₃(s) + 2Al(s) → 2Fe(s) + Al₂O₃(s).

ΔHrxn = -850 kJ; enthalpy of thermite reaction.

n(Al) = 6.00 mol; amount of the aluminium.

From the balanced chemical reaction make proportion:

2 mol : -850 kJ = 6 mol : ΔH.

ΔH = -850 kJ · 6 mol ÷ 2 mol.

ΔH = -2550 kJ; enthalpy from 6 moles of aluminium.

the answer is 75.0%.