Chapter 11, Problem 54A

(a)

Interpretation Introduction

Interpretation:

The balanced equation needs to be determined for the complete combustion of compound octane ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ assuming that the products are carbon dioxide and water.

Concept Introduction:

A hydrocarbon undergoes combustion in the presence of oxygen to produce carbon dioxide and water.

Answer to Problem 54A

The balanced equation is ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}+12.5{\text{O}}_2\to 8{\text{CO}}_{\text{2}}+9{\text{H}}_{\text{2}}\text{O}$ .

Explanation of Solution

Step 1:

It is known that when reactants are hydrocarbons in a combustion reaction, the products will be water and carbon dioxide. Thus, the equation becomes as ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$ .

Further, according to the law of conservation of mass, the mass of the product must be equal to the mass of reactants. So, on the right side of the equation ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$ , there is 1 C and on the left side, there are 8 C so one must put a coefficient of 8 in front of ${\text{CO}}_{\text{2}}$ .

Thus, the equation becomes:

  ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}+{\text{O}}_2\to 8{\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

Step 2:

Now, further on the left side of the equation ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}+{\text{O}}_2\to 8{\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$ , there are 18 H and on the right side, there are 2 H. So, one must put a coefficient of 9 in front of ${\text{H}}_{\text{2}}\text{O}$ and 12.5 in front of ${\text{O}}_2$ for making the balanced equation.

Thus, the equation becomes now as:

  ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}+12.5{\text{O}}_2\to 8{\text{CO}}_{\text{2}}+9{\text{H}}_{\text{2}}\text{O}$

Hence, the balanced chemical equation is:

  ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}+12.5{\text{O}}_2\to 8{\text{CO}}_{\text{2}}+9{\text{H}}_{\text{2}}\text{O}$

Or,

  ${\text{2C}}_{\text{8}}{\text{H}}_{\text{18}}+25{\text{O}}_2\to 16{\text{CO}}_{\text{2}}+18{\text{H}}_{\text{2}}\text{O}$

(b)

Interpretation Introduction

Interpretation:

The balanced equation needs to be determined for the complete combustion of compound glucose ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ assuming that the products are carbon dioxide and water.

Concept Introduction:

Combustion of a compound takes place in the presence of oxygen gas to produce carbon dioxide and water.

Answer to Problem 54A

The balanced equation is ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}+6{\text{O}}_2\to 6{\text{CO}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}$ .

Explanation of Solution

Step 1:

The given reaction is as follows:

  ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$ .

Further according to the law of conservation of mass, the mass of the product must be equal to the mass of reactants. So, on the right side of the equation ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$ , there is 1 C and on the left side, there are 6 C so one must put a coefficient of 6 in front of ${\text{CO}}_{\text{2}}$ .

Thus, the equation becomes:

  ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}+{\text{O}}_2\to 6{\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

Step 2:

Now, further on the left side of the equation ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}+{\text{O}}_2\to 6{\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$ , there are 12 H and on the right side, there are 2 H. So, one must put a coefficient of 6 in front of ${\text{H}}_{\text{2}}\text{O}$ and 6 in front of ${\text{O}}_2$ for making the balanced equation.

Thus, the equation becomes:

  ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}+6{\text{O}}_2\to 6{\text{CO}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}$

Hence, the balanced chemical equation is ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}+6{\text{O}}_2\to 6{\text{CO}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}$ .

(c)

Interpretation Introduction

Interpretation:

The balanced equation needs to be determined for the complete combustion of the compound ethanoic acid ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ assuming that the products are carbon dioxide and water.

Concept Introduction:

Combustion of a compound takes place in the presence of oxygen gas to produce carbon dioxide and water.

Answer to Problem 54A

The balanced equation is ${\text{HC}}_2{\text{H}}_3{\text{O}}_2+2{\text{O}}_2\to 2{\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$ .

Explanation of Solution

Step 1:

The given reaction can be represented as follows:

  ${\text{HC}}_2{\text{H}}_3{\text{O}}_2+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$ .

Further according to the law of conservation of mass, the mass of the product must be equal to the mass of reactants. So, on the right side of the equation ${\text{HC}}_2{\text{H}}_3{\text{O}}_2+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$ , there are 4 H and on the left side, there are 2 H. So, one must put a coefficient of 2 in front of ${\text{H}}_{\text{2}}\text{O}$ .

Thus, the equation becomes:

  ${\text{HC}}_2{\text{H}}_3{\text{O}}_2+{\text{O}}_2\to {\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$

Step 2:

Now, further on the left side of the equation ${\text{HC}}_2{\text{H}}_3{\text{O}}_2+{\text{O}}_2\to {\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$ , there are 2 C and on the right side, there is 1 C. So, one must put a coefficient of 2 in front of ${\text{CO}}_{\text{2}}$ and 2 in front of ${\text{O}}_2$ for making the balanced equation.

Thus, the equation becomes:

  ${\text{HC}}_2{\text{H}}_3{\text{O}}_2+2{\text{O}}_2\to 2{\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$

Hence, the balanced chemical equation is ${\text{HC}}_2{\text{H}}_3{\text{O}}_2+2{\text{O}}_2\to 2{\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$ .