To calculate the minimum amount of heat energy required to vaporize 50 grams of water at its boiling point, we can use the formula:q = n * Hvapwhere q is the heat energy required, n is the number of moles of water, and Hvap is the enthalpy of vaporization.First, we need to convert the mass of water 50 grams to moles. The molar mass of water is 18.015 g/mol, so:n = mass / molar massn = 50 g / 18.015 g/moln 2.776 molesNow, we can use the formula to calculate the heat energy required:q = n * Hvapq = 2.776 moles * 40.7 kJ/molq 112.98 kJTherefore, the minimum amount of heat energy required to vaporize 50 grams of water at its boiling point is approximately 112.98 kJ.