Given ∆Hºrxn = -1670 kJ/mol for 2Al(s) + (3/2)O2(g) --> Al2O3(s), determine ∆Hº for the reaction 2Al2O3(s) --> 4Al(s) + 3O2(g).

Question

I flipped Al(s) + (3/2)O2(g) --> Al2O3(s) to
Al2O3(s) --> A(g) + (3/2)O2(s) and multiplied the molar values by 2.

I ended up with ∆Hº = 3340 kJ/mol

Is this correct?

Answer 1

Right except for the typo. It's Al and not A in one of the equations.