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The molar heat of vaporization for methane, CH4, is 8.53 kJ/mol. How much energy is absorbed when 54.8 g of methane vaporizes at its boiling point?

6.42 kJ
29.1 kJ
137 kJ
467 kJ

Respuesta :

The answer to your question would be the second one. 29.1

Answer : The correct option is, 29.1 KJ

Explanation : Given,

Molar heat of vaporization for methane = 8.53 KJ/mole

Molar mass of methane = 16 g/mole

First we have to calculate the moles of methane.

[tex]\text{Moles of methane}=\frac{\text{Mass of methane}}{\text{Molar mass of methane}}=\frac{54.8g}{16g/mole}=3.425mole[/tex]

Now we have to calculate the amount of energy absorbed.

As, 1 mole of methane absorbs heat of vaporization = 8.53 KJ

So, 3.425 mole of methane absorbs heat of vaporization = 3.425 × 8.53 KJ = 29.1 KJ

Therefore, the amount of energy absorbed are, 29.1 KJ

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