How much energy is released when 0.40 mol c6h6(g) completely reacts with oxygen?

2c6h6(g) + 15o2(g) → 12co2(g) + 6h2o(g)

 This question asksyou to apply Hess's law. 
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C. 
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3 
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12 
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation. 
Then, 
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ 
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ 
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ 
... 
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ 
I hope this helps and my answer is right.