Chapter 5, Problem 5.7E

Balance the following equations:

a. ${\text{Li}}_{\text{3}}\text{N}\left(\text{s}\right)\to \text{Li}\left(\text{s}\right)+{\text{N}}_2\left(\text{g}\right)$

b. ${\text{Cu}}_{\text{2}}\text{O}\left(\text{s}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\to \text{CuO}\left(\text{s}\right)$

c. $\text{Zn}\left(\text{s}\right)+{\text{HNO}}_3\left(\text{aq}\right)\to \text{Zn}{\left({\text{NO}}_3\right)}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)$

d. ${\text{K}}_{\text{2}}{\text{SO}}_4\left(\text{aq}\right)+{\text{BaCl}}_{\text{2}}\left(\text{aq}\right)\to {\text{BaSO}}_4\left(\text{s}\right)+\text{KCl}\left(\text{aq}\right)$

e. $\text{dinitrogen}\text{monoxide}\to \text{nitrogen}+\text{oxygen}$

f. $\text{dinitrogen}\text{pentoxide}\to \text{nitrogen}\text{dioxide}+\text{oxygen}$

g. ${\text{P}}_4{\text{O}}_{10}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{H}}_{\text{4}}{\text{P}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{aq}\right)$

h. ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)+\text{HCl}\left(\text{aq}\right)\to {\text{CaCl}}_{\text{2}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$

Interpretation Introduction

(a)

Interpretation:

The given equation is to be balanced.

Concept introduction:

A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on the both sides of the equation is equal.

Answer to Problem 5.7E

The balanced equation is ${\text{2Li}}_{\text{3}}\text{N}\left(\text{s}\right)\to 6\text{Li}\left(\text{s}\right)+{\text{N}}_2\left(\text{g}\right)$.

Explanation of Solution

The given reaction is shown below.

${\text{Li}}_{\text{3}}\text{N}\left(\text{s}\right)\to \text{Li}\left(\text{s}\right)+{\text{N}}_2\left(\text{g}\right)$

The given reaction contains three atoms of lithium and one atom of nitrogen on the left hand side. While on the right hand side, it contains one atom of lithium and two atoms of nitrogen. Therefore, the coefficient of $2$ must be placed before ${\text{Li}}_{\text{3}}\text{N}$ and $6$ must be placed before $\text{Li}$ to get an overall balanced chemical equation. The balanced chemical reaction is shown below.

${\text{2Li}}_{\text{3}}\text{N}\left(\text{s}\right)\to 6\text{Li}\left(\text{s}\right)+{\text{N}}_2\left(\text{g}\right)$

Conclusion

The balanced equation is ${\text{2Li}}_{\text{3}}\text{N}\left(\text{s}\right)\to 6\text{Li}\left(\text{s}\right)+{\text{N}}_2\left(\text{g}\right)$.

Interpretation Introduction

(b)

Interpretation:

The given equation is to be balanced.

Concept introduction:

A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on the both sides of the equation is equal.

Answer to Problem 5.7E

The balanced equation is ${\text{4Cu}}_{\text{2}}\text{O}\left(\text{s}\right)+2{\text{O}}_{\text{2}}\left(\text{g}\right)\to 8\text{CuO}\left(\text{s}\right)$.

Explanation of Solution

The given reaction is shown below.

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

The given reaction contains two atoms of copper and three atoms of oxygen on the left hand side. While on the right hand side, it contains one atom of copper and one atom of oxygen. Therefore, the coefficient of $2$ must be placed before ${\text{O}}_{\text{2}}$, $4$ must be placed before ${\text{Cu}}_{\text{2}}\text{O}$ and $8$ must be placed before $\text{CuO}$ to get an overall balanced chemical equation. The balanced chemical reaction is shown below.

${\text{4Cu}}_{\text{2}}\text{O}\left(\text{s}\right)+2{\text{O}}_{\text{2}}\left(\text{g}\right)\to 8\text{CuO}\left(\text{s}\right)$

Conclusion

The balanced equation is ${\text{4Cu}}_{\text{2}}\text{O}\left(\text{s}\right)+2{\text{O}}_{\text{2}}\left(\text{g}\right)\to 8\text{CuO}\left(\text{s}\right)$.

Interpretation Introduction

(c)

Interpretation:

The given equation is to be balanced.

Concept introduction:

A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on the both sides of the equation is equal.

Answer to Problem 5.7E

The balanced equation is $\text{2Zn}\left(\text{s}\right)+4{\text{HNO}}_3\left(\text{aq}\right)\to 2\text{Zn}{\left({\text{NO}}_3\right)}_2\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\left(\text{g}\right)$.

Explanation of Solution

The given reaction is shown below.

$\text{Zn}\left(\text{s}\right)+{\text{HNO}}_3\left(\text{aq}\right)\to \text{Zn}{\left({\text{NO}}_3\right)}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)$

The given reaction contains one atom of zinc, one atom of hydrogen, one atom of nitrogen and three atoms of oxygen on the left side. While on the right hand side, it contains one atom of zinc, two atoms of hydrogen, two atom of nitrogen and six atoms of oxygen. Therefore, the coefficient of $2$ must be placed before $\text{Zn}$, $4$ must be placed before ${\text{HNO}}_3$, $2$ must be placed before $\text{Zn}{\left({\text{NO}}_3\right)}_2$ and $2$ must be placed before ${\text{H}}_2$ to get an overall balanced chemical equation. The balanced chemical reaction is shown below.

$\text{2Zn}\left(\text{s}\right)+4{\text{HNO}}_3\left(\text{aq}\right)\to 2\text{Zn}{\left({\text{NO}}_3\right)}_2\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\left(\text{g}\right)$

Conclusion

The balanced equation is $\text{2Zn}\left(\text{s}\right)+4{\text{HNO}}_3\left(\text{aq}\right)\to 2\text{Zn}{\left({\text{NO}}_3\right)}_2\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\left(\text{g}\right)$.

Interpretation Introduction

(d)

Interpretation:

The given equation is to be balanced.

Concept introduction:

A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on the both sides of the equation is equal.

Answer to Problem 5.7E

The balance equation is ${\text{K}}_{\text{2}}{\text{SO}}_4\left(\text{aq}\right)+{\text{BaCl}}_{\text{2}}\left(\text{aq}\right)\to {\text{BaSO}}_4\left(\text{s}\right)+2\text{KCl}\left(\text{aq}\right)$.

Explanation of Solution

The given reaction is shown below.

${\text{K}}_{\text{2}}{\text{SO}}_4\left(\text{aq}\right)+{\text{BaCl}}_{\text{2}}\left(\text{aq}\right)\to {\text{BaSO}}_4\left(\text{s}\right)+\text{KCl}\left(\text{aq}\right)$

The given reaction contains two atoms of potassium, one atom of sulfur, four atoms of oxygen, one atom barium and two atoms of chlorine on the left hand side. While on the right hand side, it contains one atom of potassium, one atom of sulfur, four atoms of oxygen, one atom barium and one atom of chlorine. Therefore, the coefficient of $2$ must be placed before $\text{KCl}$ to get an overall balanced chemical equation. The balanced chemical reaction is shown below.

${\text{K}}_{\text{2}}{\text{SO}}_4\left(\text{aq}\right)+{\text{BaCl}}_{\text{2}}\left(\text{aq}\right)\to {\text{BaSO}}_4\left(\text{s}\right)+2\text{KCl}\left(\text{aq}\right)$

Conclusion

The balanced equation is ${\text{K}}_{\text{2}}{\text{SO}}_4\left(\text{aq}\right)+{\text{BaCl}}_{\text{2}}\left(\text{aq}\right)\to {\text{BaSO}}_4\left(\text{s}\right)+2\text{KCl}\left(\text{aq}\right)$.

Interpretation Introduction

(e)

Interpretation:

The given equation is to be balanced.

Concept introduction:

A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on the both sides of the equation is equal.

Answer to Problem 5.7E

The balanced equation is ${\text{2N}}_{\text{2}}\text{O}\left(\text{g}\right)\to 2{\text{N}}_{\text{2}}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)$.

Explanation of Solution

The given reaction is shown below.

$\text{dinitrogen}\text{monoxide}\to \text{nitrogen}+\text{oxygen}$

The given reaction can be written as follows:

${\text{N}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{N}}_{\text{2}}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)$

The given reaction contains two atoms of nitrogen and one atom of oxygen on the left hand side. While on the right hand side, it contains two atoms of nitrogen and two atoms of oxygen. Therefore, the coefficient of $2$ must be placed before ${\text{N}}_{\text{2}}\text{O}$ and $2$ must be placed before ${\text{N}}_{\text{2}}$ to get an overall balanced chemical equation. The balanced chemical reaction is shown below.

${\text{2N}}_{\text{2}}\text{O}\left(\text{g}\right)\to 2{\text{N}}_{\text{2}}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)$

Conclusion

The balance equation is ${\text{2N}}_{\text{2}}\text{O}\left(\text{g}\right)\to 2{\text{N}}_{\text{2}}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)$.

Interpretation Introduction

(f)

Interpretation:

The given equation is to be balanced.

Concept introduction:

A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on the both sides of the equation is equal.

Answer to Problem 5.7E

The balanced equation is ${\text{2N}}_{\text{2}}{\text{O}}_5\left(\text{g}\right)\to 4{\text{NO}}_{\text{2}}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)$.

Explanation of Solution

The given reaction is shown below.

$\text{dinitrogen}\text{pentoxide}\to \text{nitrogen}\text{dioxide}+\text{oxygen}$

The given reaction can be written as follows:

${\text{N}}_{\text{2}}{\text{O}}_5\left(\text{g}\right)\to {\text{NO}}_{\text{2}}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)$

The given reaction contains two atoms of nitrogen and five atoms of oxygen on the left hand side. While on the right hand side, it contains one atom of nitrogen and four atoms of oxygen. Therefore, the coefficient of $2$ must be placed before ${\text{N}}_{\text{2}}{\text{O}}_5$ and $4$ must be placed before ${\text{NO}}_{\text{2}}$ to get an overall balanced chemical equation. The balanced chemical reaction is shown below.

${\text{2N}}_{\text{2}}{\text{O}}_5\left(\text{g}\right)\to 4{\text{NO}}_{\text{2}}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)$

Conclusion

The balanced equation is ${\text{2N}}_{\text{2}}{\text{O}}_5\left(\text{g}\right)\to 4{\text{NO}}_{\text{2}}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)$.

Interpretation Introduction

(g)

Interpretation:

The given equation is to be balanced.

Concept introduction:

A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on the both sides of the equation is equal.

Answer to Problem 5.7E

The balance equation is ${\text{P}}_4{\text{O}}_{10}\left(\text{s}\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{H}}_{\text{4}}{\text{P}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{aq}\right)$.

Explanation of Solution

The given reaction is shown below.

${\text{P}}_4{\text{O}}_{10}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{H}}_{\text{4}}{\text{P}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{aq}\right)$

The given reaction contains four atoms of phosphorus, eleven atoms of oxygen and two atoms of hydrogen on the left hand side. While on the right hand side, it contains two atoms of phosphorus, seven atoms of oxygen and four atoms of hydrogen. Therefore, the coefficient of $2$ must be placed before ${\text{H}}_{\text{4}}{\text{P}}_{\text{2}}{\text{O}}_{\text{7}}$ and $4$ must be placed before ${\text{H}}_{\text{2}}\text{O}$ to get an overall balanced chemical equation. The balanced chemical reaction is shown below.

${\text{P}}_4{\text{O}}_{10}\left(\text{s}\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{H}}_{\text{4}}{\text{P}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{aq}\right)$

Conclusion

The balanced equation is ${\text{P}}_4{\text{O}}_{10}\left(\text{s}\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{H}}_{\text{4}}{\text{P}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{aq}\right)$.

Interpretation Introduction

(h)

Interpretation:

The given equation is to be balanced.

Concept introduction:

A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on the both sides of the equation is equal.

Answer to Problem 5.7E

The balanced equation is ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)+2\text{HCl}\left(\text{aq}\right)\to {\text{CaCl}}_{\text{2}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$.

Explanation of Solution

The given reaction is shown below.

${\text{CaCO}}_{\text{3}}\left(\text{s}\right)+\text{HCl}\left(\text{aq}\right)\to {\text{CaCl}}_{\text{2}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$

The given reaction contains one atom of calcium, one atom of carbon, three atoms of oxygen, one atom of hydrogen and one atom of chlorine on the left hand side. While on the right hand side, it contains one atom of calcium, one atom of carbon, three atoms of oxygen, two atom of hydrogen and two atom of chlorine. Therefore, the coefficient of $2$ must be placed before $\text{HCl}$ to get an overall balanced chemical equation. The balanced chemical reaction is shown below.

${\text{CaCO}}_{\text{3}}\left(\text{s}\right)+2\text{HCl}\left(\text{aq}\right)\to {\text{CaCl}}_{\text{2}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$

Conclusion

The balanced equation is ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)+2\text{HCl}\left(\text{aq}\right)\to {\text{CaCl}}_{\text{2}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$.