Use the mole concept to determine the energy needed to vaporize 158 g of butane

In this case, we need to calculate for the molar mass of C_{4}H_{10 }to properly convert mass to kJ needed.

The **molar mass of C _{4}H_{10} **will appear as:

How much energy (kJ) is required to vaporize 158 g of butane (C_{4}H_{1}_{0}) at its boiling point, if Δ_{vap} is 24.3 kJ/mole?

(a) 15.1 kJ

(b) 66.1 kJ

(c) 89.4 kJ

(d) 11.2 kJ

(e) 38.4 kJ

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