Chapter U5.10, Problem 5E

Interpretation Introduction

Interpretation:

Balanced chemical equations must be written for the given fuels. A generalization must be made for the combustion of hydrocarbons.

Concept Introduction:

Fuels react with oxygen to produce the combusted products along with heat energy.

Hydrocarbon fuels are those which contain carbon and hydrogen atoms.

Answer to Problem 5E

Balanced chemical equations are given below.

  $\begin{array}{l}{\text{H}}_{\text{2}}\text{(g)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}{\text{(g)=H}}_{\text{2}}\text{O(l)}\\ {\text{C(g)+O}}_{\text{2}}{\text{(g)=CO}}_{\text{2}}\text{(g)}\\ {\text{CH}}_{\text{4}}{\text{(g)+2O}}_{\text{2}}{\text{(g)=CO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(l)}\\ {\text{C}}_{\text{4}}{\text{H}}_{10}\text{(g)+}\frac{13}{2}{\text{O}}_{\text{2}}{\text{(g)=4CO}}_{\text{2}}{\text{(g)+5H}}_{\text{2}}\text{O(l)}\\ {\text{C}}_6{\text{H}}_{14}\text{(g)+}\frac{19}{2}{\text{O}}_{\text{2}}{\text{(g)=6CO}}_{\text{2}}{\text{(g)+7H}}_{\text{2}}\text{O(l)}\\ {\text{C}}_8{\text{H}}_{18}\text{(g)+}\frac{25}{2}{\text{O}}_{\text{2}}{\text{(g)=8CO}}_{\text{2}}{\text{(g)+9H}}_{\text{2}}\text{O(l)}\end{array}$

  $\begin{array}{l}{\text{CH}}_3\text{OH(g)+}\frac{3}{2}{\text{O}}_{\text{2}}{\text{(g)=CO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(l)}\\ {\text{CH}}_3{\text{CH}}_{\text{2}}\text{OH(g)+}3{\text{O}}_{\text{2}}{\text{(g)=2CO}}_{\text{2}}{\text{(g)+3H}}_{\text{2}}\text{O(l)}\\ {\text{CH}}_3{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{OH(g)+6O}}_{\text{2}}{\text{(g)=4CO}}_{\text{2}}{\text{(g)+5H}}_{\text{2}}\text{O(l)}\\ {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{(g)+6O}}_{\text{2}}{\text{(g)=6CO}}_{\text{2}}{\text{(g)+6H}}_{\text{2}}\text{O(l)}\text{.}\end{array}$

Explanation of Solution

For combustion of hydrocarbon, following general equation is followed.

  ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\text{(g)+(x+}\frac{\text{y}}{\text{4}}{\text{)O}}_{\text{2}}{\text{(g)=xCO}}_{\text{2}}\text{(g)+}\frac{\text{y}}{\text{2}}{\text{H}}_{\text{2}}\text{O(l)}\text{.}$

Here x and y are the number of C and H atoms respectively.

For fuels with C, H and O like alcohols and glucose the following general equation is followed.

  ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}{\text{O}}_{\text{z}}\text{(g)+(x+}\frac{\text{y}}{\text{4}}\text{-}\frac{z}{2}{\text{)O}}_{\text{2}}{\text{(g)=xCO}}_{\text{2}}\text{(g)+}\frac{\text{y}}{\text{2}}{\text{H}}_{\text{2}}\text{O(l)}\text{.}$

x, y and z are number of C, H and O atoms respectively.

Higher the number of carbon in a hydrocarbon fuel higher is the amount of oxygen required for complete combustion.

Conclusion

Balanced equations are given for the combustion of fuels and a generalization is made for the amount of oxygen required for the combustion of hydrocarbon fuels.