#### 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=
$3.27\,mol\times22.44\,kJ/mol=73.4\,kJ$

Published by
Cengage Learning

ISBN 10:
1285199049

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

73.4 kJ

You can help us out by revising, improving and updating this answer.

Update this answerAfter you claim an answer you’ll have **24 hours** to send in a draft. An editor
will review the submission and either publish your submission or provide feedback.