Question

1. How many grams of ethanol, C2H5OH, can be boiled with 863.7 kJ of heat energy? The molar heat of vaporization of ethanol is 38.6 kJ/mol. Answer in g.

2. How much heat energy is required to boil 12.7 g of ammonia, NH3? The molar heat of vaporization of ammonia is 23.4kJ/mol. Answer in kJ

Answer 1

(1) Molar mass of ethanol, C₂H₅OH is = (2xatomic mass of C) +(6x atomic mass of H ) + (1xatomic mass of O)

                                                   = ( 2x12)+(6x1)+(1x16)

                                                  = 46 g/mol

Given the molar heat of vaporization of ethanol is 38.6 kJ/mol.

So 46 g of ethanol requires 38.6 KJ of heat to boil

     Z g of ethanol requires 863.7 KJ of heat to boil

    

(2) Molar mass of ammonia, NH3 is = (1xatomic mass of N) +(3x atomic mass of H )

                                                    = ( 1x14)+(3x1)

                                                 = 17 g/mol

Given the molar heat of vaporization of ammonia is 23.4 kJ/mol.

So 17 g of ammonial requires 23.4 KJ of heat to boil

     12.7 g of ethanol requires M KJ of heat to boil