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: 48b


$2CH_{2}$$=CHCH_{3}$ + $9O_{2}$ --> $6CO_{2}$ + $6H_{2}O$

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 9 oxygen atoms on the right-hand side, which means we need to put 4.5 in front of oxygen. 4.5 $\times$2=9 Therefore to get a whole number we balance it by adding 2 in front of the alkene. thus, now 1. There are 6 Carbon atoms on the left-hand side which gets balanced when we balance carbon dioxide with a 6. $2CH_{2}$$=CHCH_{3}$ --> $6CO_{2}$ 2. There are 12 hydrogen atoms on the left-hand side which gets balanced when we place 6 in front of a water molecule. (6$\times$2= 12) $2CH_{2}$$=CHCH_{3}$ --> $6H_{2}O$ 3. There are 18 oxygen atoms on the right-hand side. 6$\times$2= 12+6=18 These get balanced when we place 9 in front of an oxygen molecule on the left-hand side. (9$\times$2=18) $2CH_{2}$$=CHCH_{3}$ + $9O_{2}$ --> $6CO_{2}$ + $6H_{2}O$.
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