Chapter 12.2, Problem 24LC

Interpretation Introduction

Interpretation: The $12$ mole ratios from the equation for the combustion of isopropyl alcohol need to be determined.

Concept introduction: Combustion reaction is a type of chemical reaction in which a substance reacts with oxygen to form carbon dioxide and water. Both the products of combustion reactions are important and have many applications.

Explanation of Solution

To write the mole ratios, first, check whether the equation is balanced or not. If not, then balance it.

The balanced equation for the combustion of isopropyl alcohol is:

  $2{\text{C}}_3{\text{H}}_7{\text{OH(l) + 9O}}_2(g)\to 6{\text{CO}}_2(g)+8{\text{H}}_2\text{O(g)}$

The $12$ mole ratios from the equation for the combustion of isopropyl alcohol are:

1. $\frac{1{\text{ mole C}}_3{\text{H}}_{\text{7}}\text{OH}}{\frac{9}{2}{\text{ mole O}}_2}$
2. $\frac{1{\text{ mole O}}_2}{\frac{2}{9}{\text{ mole C}}_3{\text{H}}_{\text{7}}\text{OH}}$
3. $\frac{1{\text{ mole CO}}_2}{\frac{1}{3}{\text{ mole C}}_3{\text{H}}_{\text{7}}\text{OH}}$
4. $\frac{1{\text{ mole H}}_2\text{O }}{\frac{1}{4}{\text{ mole C}}_3{\text{H}}_{\text{7}}\text{OH}}$
5. $\frac{1{\text{ mole H}}_2\text{O }}{\frac{9}{8}{\text{ mole O}}_2}$
6. $\frac{1{\text{ mole CO}}_2}{\frac{3}{2}{\text{ mole O}}_2}$
7. $\frac{1{\text{ mole C}}_3{\text{H}}_7\text{OH }}{{\text{3 mole CO}}_2}$
8. $\frac{1{\text{ mole C}}_3{\text{H}}_7\text{OH }}{{\text{4 mole H}}_2\text{O}}$
9. $\frac{1{\text{ mole O}}_2}{\frac{8}{9}{\text{ mole H}}_2\text{O}}$
10. $\frac{1{\text{ mole O}}_2}{\frac{2}{3}{\text{ mole CO}}_2}$
11. $\frac{1{\text{ mole CO}}_2}{\frac{4}{3}{\text{ mole H}}_2\text{O}}$
12. $\frac{1{\text{ mole H}}_2\text{O }}{\frac{4}{3}{\text{ mole CO}}_2}$