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

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Thermochemical Equations concept. You can view video lessons to learn Thermochemical Equations. Or if you need more Thermochemical Equations practice, you can also practice Thermochemical Equations practice problems.

What is the difficulty of this problem?

Our tutors rated the difficulty of*How much energy (kJ) is required to vaporize 158 g of butane...*as medium difficulty.