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: 48a


$2CH_{3}$$CH_{2}$$CH_{2}$$CH_{3}$ + $13O_{2}$ --> $8CO_{2}$ + $10H_{2}O$

Work Step by Step

Hydrocarbons undergo a combustion reaction, Carbon dioxide and water are given off as the end products of the reaction. If we try to balance this equation directly, we get 13 oxygen atoms on the right-hand side, which means we need to put 6.5 in front of oxygen. 6.5 $\times$2=13 . Therefore to get a whole number we balance it by adding 2 in front of the alkane. thus, now 1. There are 8 Carbon atoms on the left-hand side which gets balanced when we balance carbon dioxide with an 8. $2CH_{3}$$CH_{2}$$CH_{2}$$CH_{3}$ --> $8CO_{2}$ 2. There are 20 hydrogen atoms on the left-hand side which gets balanced when we place 10 in front of a water molecule. (10$\times$2= 20) $2CH_{3}$$CH_{2}$$CH_{2}$$CH_{3}$ --> $10H_{2}O$ 3. There are 26 oxygen atoms on the right-hand side. 8$\times$2= 16+10=26 These get balanced when we place 13 in front of an oxygen molecule on the left-hand side. (13$\times$2=26) $2CH_{3}$$CH_{2}$$CH_{2}$$CH_{3}$ + $13O_{2}$ --> $8CO_{2}$ + $10H_{2}O$
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.