What is δhf° for alcl3(s)?

given: 2al(s) + 3 cl2(g) → 2 alcl3(s)
δh°rxn = - 706 kj

a) 109 kj/mol
b) 218 kj/mol
c) -353 kj/mol
d) -706 kj/mol
e) -218 kj/mol

*i think it’s c cause the standard enthalpies of formation for al and cl2 = 0 and
δh°rxn = products - reactants.

so, (2mol x δh°f alcl3(s)) - [(2mol x 0kj/mol) + (3mol x 0kj/mol)] = -706 kj

2mol x δh°f alcl3(s) = -706kj

δh°f alcl3(s) = -353kj/mol, right?

but when i googled the standard enthalpy of formation for alcl3(s), the table said it was about -704kj/mol, which means d is correct?