Question

Balance each of the following chemical equations.

1. ${\mathbf{KO}}_{\mathbf{2}}\left(\mathbf{s}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{l}\right)\to \mathbf{KOH}\left(\mathbf{aq}\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(\mathbf{g}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\left(\mathbf{aq}\right)$
2. ${\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\left(\mathbf{s}\right)\mathbf{+}{\mathbf{HNO}}_{\mathbf{3}}\left(\mathbf{aq}\right)\to \mathbf{Fe}{\left({\mathbf{NO}}_{\mathbf{3}}\right)}_{\mathbf{3}}\left(\mathbf{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{l}\right)$
3. ${\mathbf{NH}}_{\mathbf{3}}\left(\mathbf{g}\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(\mathbf{g}\right)\to \mathbf{NO}\left(\mathbf{g}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{g}\right)$
4. ${\mathbf{PCl}}_{\mathbf{5}}\left(\mathbf{l}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{l}\right)\to {\mathbf{H}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}\left(\mathbf{aq}\right)\mathbf{+}\mathbf{HCl}\left(\mathbf{g}\right)$
5. $\mathbf{CaO}\left(\mathbf{s}\right)\mathbf{+}\mathbf{C}\left(\mathbf{s}\right)\to {\mathbf{CaC}}_{\mathbf{2}}\left(\mathbf{s}\right)\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\left(\mathbf{g}\right)$
6. ${\mathbf{MoS}}_{\mathbf{2}}\left(\mathbf{s}\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(\mathbf{g}\right)\to {\mathbf{MoO}}_{\mathbf{3}}\left(\mathbf{s}\right)\mathbf{+}{\mathbf{SO}}_{\mathbf{2}}\left(\mathbf{g}\right)$

Short answer

Balanced chemicalequations are:

1. $2K{O}_2\left(s\right)+2H_{2}O\left(l\right)\to 2KOH\left(aq\right)+O_2\left(g\right)+H_2{O}_2\left(aq\right)$
2. $F{e}_2{O}_3\left(s\right)+6HN{O}_3\left(aq\right)\to 2Fe{\left(N{O}_3\right)}_3\left(aq\right)+3H_{2}O\left(l\right)$
3. $4N{H}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right)$
4. $PC{l}_5\left(l\right)+4H_{2}O\left(l\right)\to H_{3}P{O}_4\left(aq\right)+5HCl\left(g\right)$
5. localid="1654855606336" $2CaO\left(s\right)+5C\left(s\right)\to 2Ca{C}_2\left(s\right)+C{O}_2\left(g\right)$
6. $2Mo{S}_2\left(s\right)+7O_2\left(g\right)\to 2Mo{O}_3\left(s\right)+4S{O}_2\left(g\right)$
   

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.

Balancing equation (a)

The given equation is

$K{O}_2\left(s\right)+H_{2}O\left(l\right)\to KOH\left(aq\right)+O_2\left(g\right)+H_2{O}_2\left(aq\right)$

On the left side, we have two $H$atoms and three on the right side. As a result, we add a 2 coefficient for $H_{2}O$and a 2 coefficient for $\text{KOH}$. Now on the left side, we have 4 $O$ atoms, and on the right side, we have 6 atoms. As a result, we give ${\text{KO}}_2$ a 2 coefficient. Both $O$ and $K$ atoms are now in balance. As a result, we have the following equation:

$2K{O}_2\left(s\right)+2H_{2}O\left(l\right)\to 2KOH\left(aq\right)+O_2\left(g\right)+H_2{O}_2\left(aq\right)$

Balancing equation (b)

The given equation is

$F{e}_2{O}_3\left(s\right)+HN{O}_3\left(aq\right)\to Fe{\left(N{O}_3\right)}_3\left(aq\right)+H_{2}O\left(l\right)$

On the left side, we have one $H$atom and two on the right side. As a result, we add a 3 coefficient for $H_{2}O$and 6 coefficients for $HNO$. Now on the left side, we have 1 $Fe$atoms, and on the right side, we have 2 atoms. As a result, we gave $Fe{\left(N{O}_3\right)}_3$a 2 coefficient. Both $Fe$ and $H$ atoms are now in balance. As a result, we have the following equation

$F{e}_2{O}_3\left(s\right)+6HN{O}_3\left(aq\right)\to 2Fe{\left(N{O}_3\right)}_3\left(aq\right)+3H_{2}O\left(l\right)$

Balancing equation (c)

The given equation is:

$N{H}_3\left(g\right)+O_2\left(g\right)\to NO\left(g\right)+H_{2}O\left(g\right)$

On the left side, we have two $H$atoms and three on the right side. As a result, we add a 6 coefficient for $H_{2}O$and 4 coefficients for $N{H}_3$. Now on the left side, we have 1 $N$ atom, and on the right side, we have 4 atoms. So, we give $NO$a 4 coefficient. Both $N$ and $H$ atoms are now in balance. As a result, we have the following equation:

$4N{H}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right)$

Balancing equation (d)

The given equation is

$PC{l}_5\left(l\right)+H_{2}O\left(l\right)\to H_{3}P{O}_4\left(aq\right)+HCl\left(g\right)$

On the left side, we have four $H$atoms and two on the right side. As a result, we add a 5 coefficient for $HCl$and 4 coefficients for $H_{2}O$. The $H$ atoms are now in balance. As a result, we have the following equation:

$PC{l}_5\left(l\right)+4H_{2}O\left(l\right)\to H_{3}P{O}_4\left(aq\right)+5HCl\left(g\right)$

Balancing equation (e)

The given equation is

$CaO\left(s\right)+C\left(s\right)\to Ca{C}_2\left(s\right)+C{O}_2\left(g\right)$

On the left side, we have three $C$atoms and one on the right side. As a result, we add a 2 coefficient for $Ca{C}_2$and 5 coefficients for $C$. Now on the left side, we have 2 $Ca$ atoms, and on the right side, we have 1 atom. So, we gave $CaO$a 2 coefficient. Both $C$ and $CaO$ atoms are now in balance. As a result, we have the following equation:

localid="1654855682231" $2CaO\left(s\right)+5C\left(s\right)\to 2Ca{C}_2\left(s\right)+C{O}_2\left(g\right)$

Balancing equation (f)

The given equation is

$Mo{S}_2\left(s\right)+O_2\left(g\right)\to Mo{O}_3\left(s\right)+S{O}_2\left(g\right)$

On the left side, we have one $S$atom and two on the right side. As a result, we add 2 coefficient for$Mo{S}_2$ and 4 coefficients for $S{O}_2$. Now on the left side, we have 2 $Mo$ atoms, and on the right side, we have 1 atom. So, we gave $Mo{O}_3$ a 2 coefficient. Then, on the left side, we have 9 $O$ atoms, and on the right side, we have 2 atoms. So, we gave $O_2$ a 7 coefficient. So,$S$ ,$O$ , and $Ca$ atoms are now in balance. As a result, we have the following equation:

$2Mo{S}_2\left(s\right)+7O_2\left(g\right)\to 2Mo{O}_3\left(s\right)+4S{O}_2\left(g\right)$