Determine the enthalpy for this reaction: al(oh)3(s)+3hcl(g)→alcl3(s)+3h2o(l)

Important information to solve the exercise :
SubstanceΔHf (kJ/mol):
HCl(g)=−92.0 kJ/mol
Al(OH)3(s)=−1277.0 kJ/mol
 H2O(l)=−285.8 kJ/mol
AlCl3(s)=−705.6 kJ/mol

Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
    reactants                         products

products- reactants:

(−705.6) + (3 x −285.8) - ( −1277.0 ) - (3 x −92.0 ) = - 10.0 kJ per mole at 25°C

The answer is b i think