#### Answer

73.4 kJ

#### Work Step by Step

Number of moles of $C_{4}H_{10}$
$=\frac{190.\,g}{58.12\,g/mol}=3.27\,mol$
22.44 kJ is needed to vaporize 1 mol of $C_{4}H_{10}$.
$\implies$ Heat transfer required q=
$3.27\,mol\times22.44\,kJ/mol=73.4\,kJ$

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Published by
Cengage Learning

ISBN 10:
1285199049

ISBN 13:
978-1-28519-904-7

73.4 kJ

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