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: 47c


$2CH=CH $+ $5O_{2}$ --> $4CO_{2}$ + $2H_{2}O$

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 5 oxygen atoms on the right-hand side, which means we need to put 2.5 in front of oxygen. 2.5 $\times$2=5 Therefore to get a whole number we balance it by adding 2 in front of the alkyne. Thus, 1. There are 4 Carbon atoms on the left-hand side which gets balanced when we balance carbon dioxide with a 4. $2CH=CH $ --> $4CO_{2}$ 2. There are 4 hydrogen atoms on the left-hand side which gets balanced when we place 2 in front of a water molecule. (2$\times$2= 4). $2CH=CH $ --> $2H_{2}O$ 3. There are 10 oxygen atoms on the right-hand side. 4$\times$2= 8+2=10 These get balanced when we place 5 in front of an oxygen molecule on the left-hand side. (5$\times$2=10) $2CH=CH $+ $5O_{2}$ --> $4CO_{2}$ + $2H_{2}O$
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