How much energy is required to vaporize 178 g of butane (C₄H₁₀)at its boiling point,if its ΔH_vap is 24.3 kJ/mol?

Question

Quizplus · Accepted Answer

First, we need to find the number of moles of butane:

n = m/MW = 178 g / 58.12 g/mol = 3.06 mol

Then, we can use the formula:

q = nΔHvap

q = (3.06 mol)(24.3 kJ/mol)
q = 74.4 kJ

Therefore, the energy required to vaporize 178 g of butane is 74.4 kJ. The best choice is B.

How Much Energy Is Required to Vaporize 178 G of Butane