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

$CH_{3}$$CH_{2}$$CH=CH_{2}$ + $6O_{2}$ --> $4CO_{2}$ + $4H_{2}O$

#### 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 4 Carbon atoms on the left-hand side which gets balanced when we balance carbon dioxide with a 4. $CH_{3}$$CH_{2}$$CH=CH_{2}$ --> $4CO_{2}$ 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=CH_{2}$ --> $4H_{2}O$ 3. There are 12 oxygen atoms on the right-hand side. (4$\times$2= 8+4=12) These get balanced when we place 6 in front of an oxygen molecule on the left-hand side. (6$\times$2=12) $CH_{3}$$CH_{2}$$CH=CH_{2}$ + $6O_{2}$ --> $4CO_{2}$ + $4H_{2}O$

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