isarussell
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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.

Respuesta :

Edufirst

Answer:

  • 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.

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