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How much energy is required to vaporize 158 g of butane (C4H10) at its boiling point,...

How much energy is required to vaporize 158 g of butane (C4H10) at its boiling point, if its ΔHvap is 24.3 kJ/mol

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Answer #1


Molar mass of C4H10,
MM = 4*MM(C) + 10*MM(H)
= 4*12.01 + 10*1.008
= 58.12 g/mol


mass(C4H10)= 158 g

use:
number of mol of C4H10,
n = mass of C4H10/molar mass of C4H10
=(158.0 g)/(58.12 g/mol)
= 2.719 mol

use:
Q = n*Hvap
= 2.719 mol * 24.3 KJ/mol
= 66.1 KJ

Answer: 66.1 KJ

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