Question

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.

Answer 1

• 13.0kJ

Step-by-step 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.