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 C4H10 to properly convert mass to kJ needed.
The molar mass of C4H10 will appear as:
How much energy (kJ) is required to vaporize 158 g of butane (C4H10) at its boiling point, if Δvap 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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