How many kilojoules of heat are needed to completely vaporize 42.8 grams of c4h10o at its boiling point given that c4h10o has a heat of vaporization (δhvap) of +26.5 kj/mol and a molar mass of 74.12 g/mol?

Answer is: 15.30 kilojoules of heat are needed to completely vaporize C₄H₁₀.
m(C₄H₁₀) = 42.8 g.
M(C₄H₁₀) = 74.12 g/mol.
n(C₄H₁₀) = m(C₄H₁₀) ÷ M(C₄H₁₀).
n(C₄H₁₀) = 42.8 g ÷ 74.12 g/mol.
n(C₄H₁₀) = 0.577 mol.
Q = n(C₄H₁₀) · ΔHvap.
Q = 0.577 mol · 26.5 kJ/mol.
Q = 15.30 kJ, heat of butane.

2.00 is the !