How much energy is required to vaporize 185 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol?

Answer is: 73.52 kJ of energy is required to vaporize butane.
m(C₄H₁₀) = 185 g.
n(C₄H₁₀) = m(C₄H₁₀) ÷ M(C₄H₁₀).
n(C₄H₁₀) = 185 g ÷ 58.12 g/mol.
n(C₄H₁₀) = 3.18 mol; amount of butane.
Hvap = 23.1 kJ/mol; the heat of vaporization for butane.
Q = Hvap · n(C₄H₁₀).
Q = 23.1 kJ/mol · 3.18 mol; energy.
Q = 73.52 kJ.

The answer to this problem is. 67