How much energy is required to vaporize 2 kg of gold? use the table below and this equation: q = mlvapor?

If the gold is present in the liquid state, you only have to determine the latent heat of vaporization, or lvap. The empirical data for gold is 330 kJ/mol.

Q = mlvap

Q = (2 kg) (1 kmol/197 kg) (1,000 mol/1 kmol)

Q = 10.15 kJ

It needs an energy of 10.15 kilojoules.