How much energy is required to vaporize 158 g of butane (C₄H₁₀)at its boiling point,if its ΔH_vap is 24.3 kJ/mol? A)15.1 kJ B)66.1 kJ C)89.4 kJ D)11.2 kJ E)38.4 kJ

Question

Quizplus · Accepted Answer

The energy required to vaporize a substance is given by the equation:Energy = nΔHvapwhere n is the number of moles of the substance and ΔHvap is the enthalpy of vaporization.To find the number of moles of butane, we divide the mass by the molar mass:n = 158 g / 58.12 g/mol = 2.72 molNow we can calculate the energy required:Energy = 2.72 mol * 24.3 kJ/mol = 66.1 kJTherefore, the correct answer is B) 66.1 kJ.

How Much Energy Is Required to Vaporize 158 G of Butane