Answer :
Final answer:
The energy required to vaporize 158 g of butane at its boiling point is 66.2 kJ.
Explanation:
The energy required to vaporize a substance can be calculated using the equation:
ΔHvap = ΔHvap (molar) × number of moles
To calculate the number of moles of butane, we need to use its molar mass, which is the sum of the atomic masses of each element in the compound. The molar mass of butane (C4H10) is 58 g/mol.
Using the given mass of 158 g, we can calculate the number of moles: 158 g / 58 g/mol = 2.724 moles.
Finally, we can calculate the energy required to vaporize 158 g of butane at its boiling point: ΔHvap = 24.3 kJ/mol × 2.724 moles = 66.2 kJ.
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