Calculate the heat of reaction, ∆H°, for the following: 3FeCl2(s) + 4H2O(g)  Fe3O4(s) + 6HCl(g) + H2(g)

562.76 kJ

Explanation:

∆H° (kJ) is the total of the heats of formation of the products minus the  total of the heats of formation of the reactants.  

NB: look up the  heats of formation of the compounds in the heats of formation reference tables, it is found at the back of most chemistry text

heats of formation of the products (i.e.  Fe₃O₄(s) , 6HCl(g) , H₂(g) ) = -1120.9 + 6(-92.3)+ 0 = -1674.7 kJ

heats of formation of the reactants (i.e.  3FeCl₂(s) , 4H₂O(g) ) =3(-423.42) + 4(-241.8) = -2237.46kJ

∆H° = -1674.7 kJ - (-2237.46kJ) =562.76 kJ