Given that 2Al(s) + 3 2 O2(g) → Al2O3(s) ΔH o rxn = −1601 kJ/mol and
2Fe(s) + 3 2 O2(g) → Fe2O3(s)
ΔH o rxn = −821 kJ/mol
calculate the standard enthalpy change for the following reaction:
2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s)

ΔH°rxn = -780kJ

Explanation:

It is possible to find enthalpy change of a reaction by the sum of analogue reactions (Hess's law).

Using the reactions:

1. 2Al(s) + 3/2O2(g) → Al2O3(s)  ΔH°rxn = −1601 kJ/mol

2. 2Fe(s) + 3/2O2(g) → Fe2O3(s)  ΔH°rxn = −821 kJ/mol

1 - 2:

2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)

ΔH°rxn = ΔH°rxn(1) - ΔH°rxn(2)

ΔH°rxn = −1601 kJ/mol - (-821 kJ/mol)

ΔH°rxn = -780kJ

it's c. elements are placed in rows based on how may valencies electrons have.

the answers would be

the first one and the second one