How much cuo would you need to react with excess aluminum to generate same amount of heat as 141 g of fe2o3 does when reacted with excess aluminum?

Refer to this page (http://www.mrbigler.com/misc/energy-of-formation.PDF) for the data of heat of formation. For elemental forms, ΔHf is 0. Then, we use Hess' law.

ΔHrxn = Σ(n×ΔHf of products) - Σ(n×ΔHf of reactants)

Calculate ΔHrxn for Fe₂O₃:
Fe₂O₃ + 2 Al --> Al₂O₃ + 2 Fe
ΔHrxn = (-1675.7 kJ/mol) - (-824.2 kJ/mol)
ΔHrxn = -851.5 kJ/mol * 141 g * 1 mol Fe₂O₃/159.69 g
ΔHrxn = -751.84 kJ

Next, we solve ΔHrxn for CuO:
3 CuO + 2 Al --> Al₂O₃ + 3 Cu
ΔHrxn = (-1675.7 kJ/mol) - (-157.3 kJ/mol)
ΔHrxn = -1518.4 kJ/mol

Then, we find the mol CuO needed:
-751.84 = n*(-1518.4)
n = 0.495 mol

Since the molar mass of CuO is 79.545 g/mol.
Mass = 0.495 mol*79.545 g/mol = 39.4 g CuO