2al(s)+3cl2(g)→2alcl3(s) δh∘ = -1408.4 kj. how much heat (in kilojoules) is released on reaction of 4.70 g of al?

             - 122.58 kJ

Solution:
According to following  equation,

                          2 Al  +  3 Cl₂    →    2 AlCl₃     δH° = -1408.4 kJ

When,
         54 g (2 mole) Al on reaction releases  =  1408.4 kJ heat
So,
                               4.70 g of Al will release  =  X kJ of Heat

Solving for X,
                      X  =  (4.70 g × 1408.4 kJ) ÷ 54 g

                      X  =  - 122.58 kJ

The mass has escaped into a gas which you can not see so the law stays true.