Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 20 - Sections 20.1-20.14 - Exercises - Problems by Topic - Page 993: 47a

Answer

$CH_{3}$$CH_{2}$$CH_{3}$ + $5O_{2}$ --> $3CO_{2}$ + $4H_{2}O$

Work Step by Step

Hydrocarbons undergo a combustion reaction, Carbon dioxide and water are given off as the end products of the reaction. 1. There are 3 Carbon atoms on the left-hand side which gets balanced when we balance carbon dioxide with a 3. $CH_{3}$$CH_{2}$$CH_{3}$ --> $3CO_{2}$ 2. There are 8 hydrogen atoms on the left-hand side which gets balanced when we place 4 in front of a water molecule. (2$\times$4= 8) $CH_{3}$$CH_{2}$$CH_{3}$ --> $4H_{2}O$ 3. There are 10 oxygen atoms on the right-hand side. (3$\times$2= 6 +4=10) These get balanced when we place 5 in front of an oxygen molecule on the left-hand side. (5$\times$2=10) $CH_{3}$$CH_{2}$$CH_{3}$ + $5O_{2}$ --> $3CO_{2}$ + $4H_{2}O$
Update this answer!

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

Update this answer

After 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.