Question

How much energy is required to vaporize 185 g of butane at its boiling point? The heat of vaporization for butane is 23.1 kJ/mol. Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer:

73.5 kJ

Step-by-step explanation:

Mass of butane = 185 g

Heat of vaporization for butane = 23.1 kJ/mol

Molar mass of butane = 58.12 g/mol

Number of moles of butane = $\frac{\text{Mass of butane}}{\text{Molar mass of butane}}=\frac{185}{58.12}=3.18\ moles$

Energy required for burning 185 g of butane = 3.18×23.1 = 73.5 kJ

∴ Energy is required to vaporize 185 g of butane at its boiling point is 73.5 kJ