Ammonia, NH3, is used as a refrigerant. At its boiling point of -33ºC, the enthalpy of vaporization of ammonia is 23.3 kJ/mol. How much heat is required to vaporize 125 g of ammonia at -33ºC?

3 answers

171kj

q=molesx heat of vaporization
The above formula is correct but you will need to change 125 g NH3 to moles to use it.
23.3X125(1 mole NH3/17.024g(molar mass of NH3)= 171kj