How much energy is required to vaporize 158 g of butane (C₄H₁₀) at its boiling point if its Δ_vapH is 24.3 kJ mol^-1? 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 = Δ_vapH × moles of substanceThe moles of butane can be calculated using its molar mass:Moles of butane = mass / molar mass= 158 g / 58.12 g/mol= 2.72 molTherefore, the energy required to vaporize 158 g of butane is:Energy = 24.3 kJ/mol × 2.72 mol= 66.1 kJSo, the correct answer is B) 66.1 kJ.

How Much Energy Is Required to Vaporize 158 G of Butane