Determine the enthalpy for this reaction: Al (OH) 3 (s) + 3HCl (g) →AlCl3 (s) + 3H2O (l)

Important information to solve the exercise:

Substance ΔHf (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