Answer:
The amount of energy required to vaporize 1.5 kg of aluminium is 16.345 GJ
Step-by-step explanation:
Given that the heat of vaporization of aluminium, ΔHvap = 294000 kJ/mol
Mass of the given aluminium = 1.5 kG = 1500 g
Number of moles of aluminium = Mass of aluminium/(Molar Mass of aluminium)
Molar Mass of aluminium = 26.98 g/mol
Therefore, we have;
Number of moles = 1500/26.98 = 55.6 moles
The energy required = The heat of vaporization of aluminium × Number of moles of aluminium
The energy required = 294000 × 55.6 = 16345441.0675 kJ = 16.345 GJ.