Chapter 3, Problem 67E

(a)

Interpretation Introduction

Interpretation: The given unbalanced reaction is to be balanced.

  

Concept introduction: Unbalanced chemical equations just reveal the species that are a part of the reaction. However, it doesn’t give any information regarding their relative amounts needed to satisfy law of mass conservation. This is reason why equations must actually be balanced. Balance reactions are characterized by balanced mass along with charge on both sides.

Answer to Problem 67E

The balanced equation is shown below.

  ${\text{2C}}_{\text{6}}{\text{H}}_{\text{6}}\left(l\right)+15{\text{O}}_{\text{2}}\left(g\right)\to 12{\text{CO}}_2\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The given unbalanced equation is shown below.

  

Figure 1

The given unbalanced equation using chemical formula of the species interpreted from the above figure is shown below.

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

In order to balance any equation, equalize the atoms of every element on reactant’s and product’s side.

The number of atoms of $\text{C}$ on the left side is $6$ . Therefore, in order to balance the $\text{C}$ atoms, add $6$ as the stoichiometric coefficient for ${\text{CO}}_2$ on right hand right. The equation obtained after doing this is,

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)\to 6{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$ .

Similarly, the number of atoms of $\text{H}$ on the left side is $6$ . Therefore, in order to balance the $\text{H}$ atoms, add $3$ as the stoichiometric coefficient for ${\text{H}}_{\text{2}}\text{O}$ on right hand right. The equation obtained after doing this is,

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)\to 6{\text{CO}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The number of atoms of $\text{O}$ atoms on the right side is now $15$ . Therefore, in order to balance the $\text{O}$ atoms, add $\frac{15}{2}$ as the stoichiometric coefficient for ${\text{O}}_{\text{2}}$ on left hand right. The equation obtained after doing this is,

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\left(l\right)+\frac{15}{2}{\text{O}}_{\text{2}}\left(g\right)\to 6{\text{CO}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(g\right)$

In order to obtain whole number stoichiometric coefficients, multiply all stoichiometric coefficients by $2$ . The equation obtained after doing this is

  ${\text{2C}}_{\text{6}}{\text{H}}_{\text{6}}\left(l\right)+15{\text{O}}_{\text{2}}\left(g\right)\to 12{\text{CO}}_2\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Since, the number of atoms of each element is equal on both sides, therefore the above equation is balanced.

(b)

Interpretation Introduction

Interpretation: The given unbalanced reaction is to be balanced.

  

Concept introduction: Unbalanced chemical equations just reveal the species that are a part of the reaction. However, it doesn’t give any information regarding their relative amounts needed to satisfy low of mass conservation. This is reason why equations must actually be balanced. Balance reactions are characterized by balanced mass along with charge on both sides.

Answer to Problem 67E

The balanced equation is shown below.

  ${\text{2C}}_4{\text{H}}_{10}\left(g\right)+13{\text{O}}_{\text{2}}\left(g\right)\to 8{\text{CO}}_2\left(g\right)+10{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The given unbalanced equation is shown below.

  

Figure 2

The given unbalanced equation using chemical formula of the species interpreted from the above figure is shown below.

  ${\text{C}}_4{\text{H}}_{10}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

In order to balance any equation, equalize the atoms of every element on reactant’s and product’s side.

The number of atoms of $\text{C}$ on the left side is $4$ . Therefore, in order to balance the $\text{C}$ atoms, add $4$ as the stoichiometric coefficient for ${\text{CO}}_2$ on right hand right. The equation obtained after doing this is,

  ${\text{C}}_4{\text{H}}_{10}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to 4{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Similarly, the number of atoms of $\text{H}$ on the left side is $10$ . Therefore, in order to balance the $\text{H}$ atoms, add $5$ as the stoichiometric coefficient for ${\text{H}}_{\text{2}}\text{O}$ on right hand right. The equation obtained after doing this is,

  ${\text{C}}_4{\text{H}}_{10}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to 4{\text{CO}}_2\left(g\right)+5{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The number of atoms of $\text{O}$ atoms on the right side is now $13$ . Therefore, in order to balance the $\text{O}$ atoms, add $\frac{13}{2}$ as the stoichiometric coefficient for ${\text{O}}_{\text{2}}$ on left hand right. The equation obtained after doing this is,

  ${\text{C}}_4{\text{H}}_{10}\left(g\right)+\frac{13}{2}{\text{O}}_{\text{2}}\left(g\right)\to 4{\text{CO}}_2\left(g\right)+5{\text{H}}_{\text{2}}\text{O}\left(g\right)$

In order to obtain whole number stoichiometric coefficients, multiply all stoichiometric coefficients by $2$ .The equation obtained after doing this is,

  ${\text{2C}}_4{\text{H}}_{10}\left(g\right)+13{\text{O}}_{\text{2}}\left(g\right)\to 8{\text{CO}}_2\left(g\right)+10{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Since, the number of atoms of each element is equal on both sides, therefore the above equation is balanced.

(c)

Interpretation Introduction

Interpretation: The reaction ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$ is to be balanced.

Concept introduction: Unbalanced chemical equations just reveal the species that are a part of the reaction. However, it doesn’t give any information regarding their relative amounts needed to satisfy low of mass conservation. This is reason why equations must actually be balanced. Balance reactions are characterized by balanced mass along with charge on both sides.

Answer to Problem 67E

The balanced equation is shown below.

  ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\left(s\right)+12{\text{O}}_{\text{2}}\left(g\right)\to 12{\text{CO}}_2\left(g\right)+11{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The given unbalanced equation is shown below.

  ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

In order to balance any equation, equalize the atoms of every element on reactant’s and product’s side.

The number of atoms of $\text{C}$ on the left side is $12$ . Therefore, in order to balance the $\text{C}$ atoms, add $12$ as the stoichiometric coefficient for ${\text{CO}}_2$ on right hand right. The equation obtained after doing this is,

  ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to 12{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Similarly, the number of atoms of $\text{H}$ on the left side is $22$ . Therefore, in order to balance the $\text{H}$ atoms, add $11$ as the stoichiometric coefficient for ${\text{H}}_{\text{2}}\text{O}$ on right hand right. The equation obtained after doing this is,

  ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to 12{\text{CO}}_2\left(g\right)+11{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The number of atoms of $\text{O}$ atoms on the right side is now $35$ . Therefore, in order to balance the $\text{O}$ atoms, add $12$ as the stoichiometric coefficient for ${\text{O}}_{\text{2}}$ on left hand right. The equation obtained after doing this is,

  ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\left(s\right)+12{\text{O}}_2\left(g\right)\to 12{\text{CO}}_2\left(g\right)+11{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Since, the number of atoms of each element is equal on both sides, therefore the above equation is balanced.

(d)

Interpretation Introduction

Interpretation: The reaction $\text{Fe}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$ is to be balanced.

Concept introduction: Unbalanced chemical equations just reveal the species that are a part of the reaction. However, it doesn’t give any information regarding their relative amounts needed to satisfy low of mass conservation. This is reason why equations must actually be balanced. Balance reactions are characterized by balanced mass along with charge on both sides.

Answer to Problem 67E

The balanced equation is shown below.

  $\text{4Fe}\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$

Explanation of Solution

The given unbalanced equation is shown below.

  $\text{Fe}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$

In order to balance any equation, equalize the atoms of every element on reactant’s and product’s side.

The number of atoms of $\text{O}$ on the left side is $2$ and on the right hand side are $3$ . Therefore, in order to balance the $\text{O}$ atoms, add $2$ as the stoichiometric coefficient for ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ and $3$ as the stoichiometric coefficient for ${\text{O}}_{\text{2}}$ . The equation obtained after doing this is,

  $\text{Fe}\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$

Similarly, the number of atoms of $\text{Fe}$ on the right side is $4$ . Therefore, in order to balance the $\text{Fe}$ atoms, add $4$ as the stoichiometric coefficient for $\text{Fe}$ on left hand right. The equation obtained after doing this is,

  $\text{4Fe}\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$

Since, the number of atoms of each element is equal on both sides, therefore the above equation is balanced.

(e)

Interpretation Introduction

Interpretation: The reaction $\text{FeO}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$ is to be balanced.

Concept introduction: Unbalanced chemical equations just reveal the species that are a part of the reaction. However, it doesn’t give any information regarding their relative amounts needed to satisfy low of mass conservation. This is reason why equations must actually be balanced. Balance reactions are characterized by balanced mass along with charge on both sides.

Answer to Problem 67E

The balanced equation is shown below.

  $\text{4FeO}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$

Explanation of Solution

The given unbalanced equation is shown below.

  $\text{FeO}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$

In order to balance any equation, equalize the atoms of every element on reactant’s and product’s side.

The number of atoms of $\text{Fe}$ on the right side is $2$ and on the left hand side are $1$ . Therefore, in order to balance the $\text{Fe}$ atoms, add $2$ as the stoichiometric coefficient for $\text{FeO}$ . The equation obtained after doing this is

  $\text{2FeO}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$

Similarly, the number of atoms of $\text{O}$ on the right side is $3$ and in the left hand side is $4$ . Therefore, in order to balance the $\text{O}$ atoms, add $2$ as the stoichiometric coefficient for ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ on right hand right and replace the stoichiometric coefficient of $\text{FeO}$ by $4$ . The equation obtained after doing this is

  $\text{4FeO}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$

Since, the number of atoms of each element is equal on both sides, therefore the above equation is balanced.