Liquid butane, C4H10, is stored in cylinders to be used

as a fuel. Suppose 35.5 g of butane gas is removed from a

cylinder. How much heat must be provided to vaporize this

much gas? The heat of vaporization of butane is 21.3 kJ/mol.

13.0kJ

Explanation:

The heat to vaporize a liquid is equal to the amount of liquid in moles multiplied by the specific heat of vaporiztion per mole.

First, calculate the number of moles in 35.5g of butane.

Molar mass of butane: 58.124 g/mol

Number of moles = mass in grams/molar mass

Number of moles = 35.5g / 58.124g/mol = 0.6107632mol

Now, calculate the heat to vaporize that amount of liquid butane:

Heat = number of moles * specific heat of vaporization

Heat = 0.6107632mol * 21.3kJ/mol = 13.0 kJ

The answer must be reported with 3 significant figures.