Answer : The energy removed must be, -67.7 kJ
Solution :
The process involved in this problem are :

The expression used will be:
![\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bm%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= heat released by the reaction = ?
m = mass of benzene = 125 g
= specific heat of gaseous benzene = 
= specific heat of liquid benzene = 
= enthalpy change for vaporization = 
Molar mass of benzene = 78.11 g/mole
Now put all the given values in the above expression, we get:
![\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B125g%5Ctimes%201.06J%2Fg.K%5Ctimes%20%28353.0-%28425.0%29%29K%5D%2B125g%5Ctimes%20-434.0J%2Fg%2B%5B125g%5Ctimes%201.73J%2Fg.K%5Ctimes%20%28335.0-353.0%29K%5D)

Therefore, the energy removed must be, -67.7 kJ