## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 4 - Exercises - Page 190: 105

#### Answer

3.1 kg of carbon dioxide are added to the atmosphere per 1.0 kg of octane burned.

#### Work Step by Step

1. This is a combustion reaction; therefore, octane will react with $O_2$ to produce carbon dioxide and water: $$C_8H_{18} + O_2 \longrightarrow CO_2 + H_2O$$ - Balance the amount of carbon: $$C_8H_{18} + O_2 \longrightarrow 8CO_2 + H_2O$$ - Balance the amount of hydrogen: $$C_8H_{18} + O_2 \longrightarrow 8CO_2 + 9H_2O$$ - Balance the amount of oxygen: $$C_8H_{18} + \frac{25}{2}O_2 \longrightarrow 8CO_2 + 9H_2O$$ - Multiply all coefficients by 2 to remove the fraction: $$2 C_8H_{18} + 25O_2 \longrightarrow 16CO_2 + 18H_2O$$ $C_8H_{18}$ : ( 1.008 $\times$ 18 )+ ( 12.01 $\times$ 8 )= 114.22 kg/kmol $$\frac{1 \space kmol\space C_8H_{18} }{ 114.22 \space kg \space C_8H_{18} } \space and \space \frac{ 114.22 \space kg \space C_8H_{18} }{1 \space kmol\space C_8H_{18} }$$ $CO_2$ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 kg/kmol $$\frac{1 \space kmol\space CO_2 }{ 44.01 \space kg \space CO_2 } \space and \space \frac{ 44.01 \space kg \space CO_2 }{1 \space kmol\space CO_2 }$$ $$1.0 \space kg \space C_8H_{18} \times \frac{1 \space kmol\space C_8H_{18} }{ 114.22 \space kg \space C_8H_{18} } \times \frac{ 16 \space kmol \space CO_2 }{ 2 \space kmol \space C_8H_{18} } \times \frac{ 44.01 \space kg \space CO_2 }{1 \space kmol\space CO_2 } = 3.1 \space kg \space CO_2$$

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