To find out how much energy is absorbed when 16.5g of liquid methane vaporizes, we need to use the molar mass of methane (CH4) and the enthalpy of vaporization.
1. Calculate the number of moles of methane:
Molar mass of CH4 = 12.01g/mol + 4(1.008g/mol) = 16.04g/mol
Number of moles of CH4 = Mass / Molar mass = 16.5g / 16.04g/mol = 1.0266 mol
2. Calculate the energy absorbed:
Energy absorbed = Number of moles * Enthalpy of vaporization
Energy absorbed = 1.0266 mol * -8.17 kJ/mol = -8.38 kJ
(Note: The negative sign indicates that energy is absorbed.)
Therefore, when 16.5g of liquid methane vaporizes, approximately 8.38 kJ of energy is absorbed.
methane (CH4) has a normal boiling point of -161.6*c at this temprature the H concentration = -8.17kj/mol. if 16.5g of liquid methane vaporize. how much energy is absorbed
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