Question

A sample of 55.0 g of ethanol at 25°C is vaporized at its normal boiling point. The enthalpy of vaporization for ethanol is 38.56 kJ/mol. Calculate the amount of heat energy required for this process.

Answer 1

To calculate the amount of heat energy required for this process, we first need to determine the number of moles of ethanol in the 55.0 g sample. The molecular weight of ethanol  C2H5OH  is: 2 * 12.01 g/mol for C  +  6 * 1.01 g/mol for H  +  1 * 16.00 g/mol for O  = 46.07 g/molNow, we can find the number of moles of ethanol:55.0 g / 46.07 g/mol = 1.194 moles of ethanolNext, we can use the enthalpy of vaporization to calculate the heat energy required:Heat energy =  moles of ethanol  *  enthalpy of vaporization Heat energy =  1.194 moles  *  38.56 kJ/mol  = 46.03 kJTherefore, the amount of heat energy required to vaporize the 55.0 g sample of ethanol at its normal boiling point is 46.03 kJ.