Chapter 20 - Sections 20.1-20.14 - Exercises - Problems by Topic - Page 993: 48c

$2CH=CCH_{2}$$CH_{3} + 11O_{2}--> 8CO_{2} + 6H_{2}O Work Step by Step Hydrocarbons undergo a combustion reaction, Carbon dioxide and water are given off as the end products of the reaction. If we try to balance this equation directly, we get 11 oxygen atoms on the right-hand side, which means we need to put 5.5 in front of oxygen. 5.5 \times2=11 Therefore to get a whole number we balance it by adding 2 in front of the alkyne. Thus, now 1. There are 8 Carbon atoms on the left-hand side which gets balanced when we balance carbon dioxide with an 8. 2CH=CCH_{2}$$CH_{3}$ --> $8CO_{2}$ 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=CCH_{2}$$CH_{3} --> 6H_{2}O 3. There are 22 oxygen atoms on the right-hand side. 8\times2= 16+6=22 These get balanced when we place 9 in front of an oxygen molecule on the left-hand side. (11\times2=22) 2CH=CCH_{2}$$CH_{3}$ + $11O_{2}$--> $8CO_{2}$ + $6H_{2}O$

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