How much heat energy is required to boil 47.4 g of ammonia, NH3? The molar heat of vaporization of ammonia is 23.4 kJ/mol.

Answer 1

If we assume that the temperature of the NH3 is already at the boiling point, then we can solve this problem as follows:

molar mass NH3 = 17 g / mol

moles NH3 = 47.4 g x 1 mol / 17 g = 2.79 mols

heat needed = 2.79 mols x 23.4 kJ / mol = 65.2 kJ of heat