Question

Balance the following equations representing combusting reactions:

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

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

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

Short answer

The balanced equationsrepresenting combusting reactions are:

1. $2C_6{H}_6\left(l\right)+15O_2\left(g\right)\to 12C{O}_2\left(g\right)+6H_{2}O\left(g\right)$
2. $2C_4{H}_{10}\left(g\right)+13O_2\left(g\right)\to 8C{O}_2\left(g\right)+10H_{2}O\left(g\right)$
3. $C_{12}{H}_{22}{O}_{11}\left(s\right)+12O_2\left(g\right)\to 12C{O}_2\left(g\right)+11H_{2}O\left(g\right)$
4. $4Fe\left(s\right)+3O_2\left(g\right)\to 2F{e}_2{O}_3\left(s\right)$
5. $4FeO\left(s\right)+4O_2\left(g\right)\to 2F{e}_2{O}_3\left(s\right)$

Step-by-step solution

Definition

A balanced chemical equation is one in which the number of atoms present on the reactant side of the equation equals the number of atoms present on the product side of the equation.

Determination of equation (a)

a)

The first reactant in the image has six carbon atoms and six hydrogen atoms, indicating that it is a $C_6{H}_6$molecule. Because the second reactant only has 2 $O$atoms, it is a $O_2$. The first of the products has one $C$atom and two $O$atoms, indicating that it is a $C{O}_2$molecule. The second product molecule is a $H_{2}O$molecule with two $H$atoms and one $O$atom. As a result, the reaction can be written as follows:

$C_6{H}_6\left(l\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+H_{2}O\left(g\right)$

Balancing reactants

The reaction reveals that the reactants have 6$C$atoms, whereas the products have only 1$C$ atom. As a result, the $C$atoms can be initially balanced by applying a coefficient of 6 to $C{O}_2$, as shown below:

$C_6{H}_6\left(l\right)+O_2\left(g\right)\to 6C{O}_2\left(g\right)+H_{2}O\left(g\right)$

Balancing Hydrogen and oxygen atoms

The hydrogen atoms can then be balanced by assigning a coefficient of 3 to $H_{2}O$, resulting in a total of 6 $H$atoms on both sides of the process, as shown below:

$C_6{H}_6\left(l\right)+O_2\left(g\right)\to 6C{O}_2\left(g\right)+3H_{2}O\left(g\right)$

Only the oxygen atoms remain imbalanced now. The products have a total of 15 atoms $O$, whereas the reactant has only two. This can be balanced by applying a $\frac{15}{2}$coefficient to $O$, as seen below:

$C_6{H}_6\left(l\right)+\frac{15}{2}{O}_2\left(g\right)\to 6C{O}_2\left(g\right)+3H_{2}O\left(g\right)$

Balanced Equation

To get whole number coefficients for all the reactants and products, multiply the following equation by two. As a result, the reaction's final complete balanced equation would be as follows:

$2C_6{H}_6\left(l\right)+15O_2\left(g\right)\to 12C{O}_2\left(g\right)+6H_{2}O\left(g\right)$

Determination of equation (b)

b)

The first reactant in the image has four carbon atoms and ten hydrogen atoms, making it a $C_4{H}_{10}$ molecule. Because the second reactant contains just 2$O$ atoms, it is a $O_2$. The first of the products has one $C$atom and two $O$atoms, indicating that it is a $C{O}_2$molecule. The second product molecule is a $H_{2}O$molecule with two $H$atoms and one$O$ atom. As a result, the reaction can be written as follows:

$C_4{H}_{10}\left(g\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+H_{2}O\left(g\right)$

Balancing reactants

The reaction reveals that the reactants have 4 $C$atoms, whereas the products have only 1 $C$ atom. As a result, the $C$atoms can be initially balanced by applying a coefficient of 4 to $C{O}_2$, as shown below:

$C_4{H}_{10}\left(g\right)+O_2\left(g\right)\to 4C{O}_2\left(g\right)+H_{2}O\left(g\right)$

Balancing Hydrogen and oxygen atoms

The hydrogen atoms can then be balanced by assigning a coefficient of 5 to $H_{2}O$ , resulting in a total of 10 $H$atoms on both sides of the process, as shown below:

$C_4{H}_{10}\left(g\right)+O_2\left(g\right)\to 4C{O}_2\left(g\right)+5H_{2}O\left(g\right)$

Only the oxygen atoms remain imbalanced now. The products have a total of 13 atoms $O$, whereas the reactant has only two. This can be balanced by applying a $\frac{13}{2}$coefficient to $O$, as seen below:

$C_4{H}_{10}\left(g\right)+\frac{13}{2}{O}_2\left(g\right)\to 4C{O}_2\left(g\right)+5H_{2}O\left(g\right)$

Balanced Equation

To get whole number coefficients for all the reactants and products, multiply the following equation by two. As a result, the reaction's final complete balanced equation would be as follows:

$2C_4{H}_{10}\left(g\right)+13O_2\left(g\right)\to 8C{O}_2\left(g\right)+10H_{2}O\left(g\right)$

Balancing carbon atoms

c)

The given equation is:

$C_{12}{H}_{22}{O}_{11}\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+H_{2}O\left(g\right)$

The reaction reveals that the reactants have 12 $C$atoms, whereas the products have only 1$C$ atom. As a result, the $C$atoms can be initially balanced by applying a coefficient of 12 to $C{O}_2$, as shown below:

$C_{12}{H}_{22}{O}_{11}\left(s\right)+O_2\left(g\right)\to 12C{O}_2\left(g\right)+H_{2}O\left(g\right)$

Balancing hydrogen atoms

The hydrogen atoms can then be balanced by assigning a coefficient of 11 to $H_{2}O$, resulting in a total of 22 $H$atoms on both sides of the process, as shown below:

$C_{12}{H}_{22}{O}_{11}\left(s\right)+O_2\left(g\right)\to 12C{O}_2\left(g\right)+11H_{2}O\left(g\right)$

Balanced Equation

Only the oxygen atoms remain imbalanced now. The products have a total of 33 atoms $O$, whereas the reactant has only 13. This can be balanced by applying a 12 coefficient to $O_2$. As a result, the reaction's final complete balanced equation would be as follows:

$C_{12}{H}_{22}{O}_{11}\left(s\right)+12O_2\left(g\right)\to 12C{O}_2\left(g\right)+11H_{2}O\left(g\right)$

Balancing Fe atoms

d)

The given equation is

$Fe\left(s\right)+O_2\left(g\right)\to F{e}_2{O}_3\left(s\right)$

The reaction reveals that the products have 2 $Fe$atoms. As a result, the $C$atoms can be initially balanced by applying a coefficient of 2 to $Fe$, as shown below:

$2Fe\left(s\right)+O_2\left(g\right)\to F{e}_2{O}_3\left(s\right)$

Balancing oxygen atoms

The products have 2 $O$atoms As a result; the $O$atoms can be balanced by applying a coefficient of $\frac{3}{2}$to $O_2$, as shown below:

$2Fe\left(s\right)+\frac{3}{2}{O}_2\left(g\right)\to F{e}_2{O}_3\left(s\right)$

Balanced Equation

To get whole number coefficients for all the reactants and products, multiply the following equation by two. As a result, the reaction's final complete balanced equation would be as follows:

$4Fe\left(s\right)+3O_2\left(g\right)\to 2F{e}_2{O}_3\left(s\right)$

Balancing Fe atoms

e)

The given equation is

$FeO\left(s\right)+O_2\left(g\right)\to F{e}_2{O}_3\left(s\right)$

The reaction reveals that the products have 2$Fe$ atoms. As a result, the $C$atoms can be initially balanced by applying a coefficient of 2 to $Fe$, as shown below:

$2FeO\left(s\right)+O_2\left(g\right)\to F{e}_2{O}_3\left(s\right)$

Balancing oxygen atoms

The products have 2 $O$atoms. As a result, the $O$ atoms can be balanced by applying a coefficient of $\frac{1}{2}$to $O_2$, as shown below:

$2FeO\left(s\right)+\frac{1}{2}{O}_2\left(g\right)\to F{e}_2{O}_3\left(s\right)$

Balanced Equation

To get whole number coefficients for all the reactants and products, multiply the following equation by two. As a result, the reaction's final complete balanced equation would be as follows:

$4FeO\left(s\right)+4O_2\left(g\right)\to 2F{e}_2{O}_3\left(s\right)$