Chapter 3, Problem 70E

(a)

Interpretation Introduction

Interpretation: The balanced form of chemical equation ${\text{ KO}}_2\left(s\right)+{\text{H}}_2\text{O}\left(l\right)\to \text{KOH}\left(aq\right)+{\text{O}}_2\left(g\right)+{\text{H}}_2{\text{O}}_2\left(aq\right)$ is to be stated.

Concept introduction: In any balanced chemical equation, the number of all the atoms or elements exists on the reactant side equal to the number of all the atoms or elements present on the product side. Every measurement is dependent on the balanced equation. This is further used in determining the number of moles of reacting species, which is the key step in every calculation.

Answer to Problem 70E

The balanced form of chemical equation is $2{\text{KO}}_2\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\to 2\text{KOH}\left(aq\right)+{\text{O}}_2\left(g\right)+{\text{H}}_2{\text{O}}_2\left(aq\right)$ .

Explanation of Solution

The given chemical equation is given below.

  ${\text{ KO}}_2\left(s\right)+{\text{H}}_2\text{O}\left(l\right)\to \text{KOH}\left(aq\right)+{\text{O}}_2\left(g\right)+{\text{H}}_2{\text{O}}_2\left(aq\right)$

The number of oxygen atom on reactant side is three whereas on product side is five.

The number of hydrogen atom on reactant side is two whereas on product side is three.

The number of potassium atom on reactant side and product side is one.

In order to balance the equation, add coefficient $2$ in front of ${\text{KO}}_2\left(s\right)$ and $\text{KOH}\left(aq\right)$ . The equation would become as follows.

  $2{\text{KO}}_2\left(s\right)+{\text{H}}_2\text{O}\left(l\right)\to 2\text{KOH}\left(aq\right)+{\text{O}}_2\left(g\right)+{\text{H}}_2{\text{O}}_2\left(aq\right)$

The number of hydrogen atom on reactant side is two whereas on product side is four.

The number of potassium atom on reactant side and product side is two.

The number of oxygen atom on reactant side is five whereas on product side is six.

In order to balance the equation, add coefficient $2$ in front of ${\text{H}}_2\text{O}\left(l\right)$ on reactant side. The equation would become as follows.

  $2{\text{KO}}_2\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\to 2\text{KOH}\left(aq\right)+{\text{O}}_2\left(g\right)+{\text{H}}_2{\text{O}}_2\left(aq\right)$

Thus, all the atoms of potassium, hydrogen and oxygen are balanced on reactant and product side.

Therefore, the balanced chemical equation is $2{\text{KO}}_2\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\to 2\text{KOH}\left(aq\right)+{\text{O}}_2\left(g\right)+{\text{H}}_2{\text{O}}_2\left(aq\right)$ .

(b)

Interpretation Introduction

Interpretation: The balanced form of chemical equation ${\text{ Fe}}_2{\text{O}}_3\left(s\right)+{\text{HNO}}_3\left(l\right)\to \text{Fe}{\left({\text{NO}}_3\right)}_3\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$ is to be stated.

Concept introduction: In any balanced chemical equation, the number of all the atoms or elements exists on the reactant side equal to the number of all the atoms or elements present on the product side. Every measurement is dependent on the balanced equation. This is further used in determining the number of moles of reacting species, which is the key step in every calculation.

Answer to Problem 70E

The balanced form of chemical equation is ${\text{ Fe}}_2{\text{O}}_3\left(s\right)+6{\text{HNO}}_3\left(l\right)\to 2\text{Fe}{\left({\text{NO}}_3\right)}_3\left(aq\right)+{\text{3H}}_2\text{O}\left(l\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{ Fe}}_2{\text{O}}_3\left(s\right)+{\text{HNO}}_3\left(l\right)\to \text{Fe}{\left({\text{NO}}_3\right)}_3\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

The number of oxygen atom on reactant side is six whereas on product side is ten.

The number of hydrogen atom on reactant side is one whereas on product side is two.

The number of iron atom on reactant side is two and product side is one.

The number of nitrogen atom on reactant side is one and product side is three.

To balance the equation, add coefficient $6$ in front of ${\text{HNO}}_3\left(l\right)$ and $2$ in front of $\text{Fe}{\left({\text{NO}}_3\right)}_3\left(aq\right)$ .

The equation would become as follows.

  ${\text{ Fe}}_2{\text{O}}_3\left(s\right)+6{\text{HNO}}_3\left(l\right)\to 2\text{Fe}{\left({\text{NO}}_3\right)}_3\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

The number of oxygen atom on reactant side is twenty-one whereas on product side is nineteen.

The number of hydrogen atom on reactant side is six whereas on product side is two.

The number of iron atom on reactant side and product side is two.

The number of nitrogen atom on reactant side and product side is six.

In order to balance the equation, add coefficient three in front of ${\text{H}}_2\text{O}\left(l\right)$ . The equation would become as follows.

  ${\text{ Fe}}_2{\text{O}}_3\left(s\right)+6{\text{HNO}}_3\left(l\right)\to 2\text{Fe}{\left({\text{NO}}_3\right)}_3\left(aq\right)+{\text{3H}}_2\text{O}\left(l\right)$

Thus, all the atoms of iron, nitrogen, hydrogen, and oxygen are balanced on reactant and product side.

Therefore, the balanced chemical equation is ${\text{ Fe}}_2{\text{O}}_3\left(s\right)+6{\text{HNO}}_3\left(l\right)\to 2\text{Fe}{\left({\text{NO}}_3\right)}_3\left(aq\right)+{\text{3H}}_2\text{O}\left(l\right)$ .

(c)

Interpretation Introduction

Interpretation: The balanced form of chemical equation ${\text{ NH}}_3\left(g\right)+{\text{O}}_2\left(g\right)\to \text{NO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$ is to be stated.

Concept introduction: In any balanced chemical equation, the number of all the atoms or elements exists on the reactant side equal to the number of all the atoms or elements present on the product side. Every measurement is dependent on the balanced equation. This is further used in determining the number of moles of reacting species, which is the key step in every calculation.

Answer to Problem 70E

The balanced form of chemical equation is $4{\text{NH}}_3\left(g\right)+{\text{5O}}_2\left(g\right)\to 4\text{NO}\left(g\right)+{\text{6H}}_2\text{O}\left(g\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{ NH}}_3\left(g\right)+{\text{O}}_2\left(g\right)\to \text{NO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

The number of oxygen atom on reactant and product side is two.

The number of hydrogen atom on reactant side is three whereas on product side is two.

The number of nitrogen atom on reactant and product side is one.

To balance the equation, add coefficient $4$ in front of ${\text{NH}}_3\left(g\right)$ and $\text{NO}\left(g\right)$ . The equation would become as follows.

  $4{\text{NH}}_3\left(g\right)+{\text{O}}_2\left(g\right)\to 4\text{NO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

The number of oxygen atom on reactant side is two whereas on product side is five.

The number of hydrogen atom on reactant side is twelve whereas on product side is two.

The number of nitrogen atom on reactant side and product side is four.

To balance the equation, add coefficient $6$ in front of ${\text{H}}_2\text{O}\left(g\right)$ .and 5 in front of ${\text{O}}_2\left(g\right)$ .

The equation would become as follows.

  $4{\text{NH}}_3\left(g\right)+{\text{5O}}_2\left(g\right)\to 4\text{NO}\left(g\right)+{\text{6H}}_2\text{O}\left(g\right)$

Thus, all the atoms of nitrogen, hydrogen, and oxygen are balanced on reactant and product side.

Therefore, the balanced chemical equation is $4{\text{NH}}_3\left(g\right)+{\text{5O}}_2\left(g\right)\to 4\text{NO}\left(g\right)+{\text{6H}}_2\text{O}\left(g\right)$.

(d)

Interpretation Introduction

Interpretation: The balanced form of chemical equation, ${\text{ PCl}}_5\left(l\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+\text{HCl}\left(g\right)$ is to be stated.

Concept introduction: In any balanced chemical equation, the number of all the atoms or elements exists on the reactant side equal to the number of all the atoms or elements present on the product side. Every measurement is dependent on the balanced equation. This is further used in determining the number of moles of reacting species, which is the key step in every calculation.

Answer to Problem 70E

The balanced form of chemical equation is ${\text{ PCl}}_5\left(l\right)+4{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+5\text{HCl}\left(g\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{ PCl}}_5\left(l\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+\text{HCl}\left(g\right)$

The number of oxygen atom on reactant side is one whereas on product side is four.

The number of hydrogen atom on reactant side is two whereas on product side is four.

The number of phosphorus atom on reactant side and product side is one.

The number of chlorine atom on reactant side is five and product side is one.

To balance the equation, add coefficient $4$ in front of ${\text{H}}_2\text{O}\left(l\right)$ and $5$ in front of $\text{HCl}\left(g\right)$ .

The equation would become as follows.

  ${\text{ PCl}}_5\left(l\right)+4{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+5\text{HCl}\left(g\right)$

Thus, all the atoms of phosphorous, oxygen, hydrogen and chlorine are balanced.

Therefore, the balanced chemical equation is ${\text{ PCl}}_5\left(l\right)+4{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+5\text{HCl}\left(g\right)$ .

(e)

Interpretation Introduction

Interpretation: The balanced form of chemical equation $\text{ CaO}\left(s\right)+\text{C}\left(s\right)\to {\text{CaC}}_{\text{2}}\left(s\right)+{\text{CO}}_2\left(g\right)$ is to be stated.

Concept introduction: In any balanced chemical equation, the number of all the atoms or elements exists on the reactant side equal to the number of all the atoms or elements present on the product side. Every measurement is dependent on the balanced equation. This is further used in determining the number of moles of reacting species, which is the key step in every calculation.

Answer to Problem 70E

The balanced form of chemical equation is $2\text{CaO}\left(s\right)+\text{5C}\left(s\right)\to 2{\text{CaC}}_{\text{2}}\left(s\right)+{\text{CO}}_2\left(g\right)$ .

Explanation of Solution

The given chemical equation is given below.

  $\text{ CaO}\left(s\right)+\text{C}\left(s\right)\to {\text{CaC}}_{\text{2}}\left(s\right)+{\text{CO}}_2\left(g\right)$

The number of oxygen atom on reactant side is one whereas on product side is two.

The number of carbon atom on reactant side is one whereas on product side is three.

The number of calcium atom on reactant side and product side is one.

To balance the equation, add coefficient $2$ in front of $\text{ CaO}\left(s\right)$ , and ${\text{CaC}}_{\text{2}}\left(s\right)$ .

The equation would become as follows.

  $\text{ 2CaO}\left(s\right)+\text{C}\left(s\right)\to 2{\text{CaC}}_{\text{2}}\left(s\right)+{\text{CO}}_2\left(g\right)$

The number of oxygen atom on reactant and product side is two.

The number of carbon atom on reactant side is one whereas on product side is five.

The number of calcium atom on reactant side and product side is two.

To balance the equation, add coefficient $5$ in front of $\text{C}\left(s\right)$ . The equation would become as follows.

  $\text{ 2CaO}\left(s\right)+5\text{C}\left(s\right)\to 2{\text{CaC}}_{\text{2}}\left(s\right)+{\text{CO}}_2\left(g\right)$

Thus, all the atoms of calcium, oxygen and carbon on reactant and product side are balanced.

Therefore, the balanced chemical equation is $\text{ 2CaO}\left(s\right)+5\text{C}\left(s\right)\to 2{\text{CaC}}_{\text{2}}\left(s\right)+{\text{CO}}_2\left(g\right)$ .

(f)

Interpretation Introduction

Interpretation: The balanced form of chemical equation ${\text{ MoS}}_2\left(s\right)+{\text{O}}_2\left(g\right)\to {\text{MoO}}_{\text{3}}\left(s\right)+{\text{SO}}_2\left(g\right)$ is to be stated.

Concept introduction: In any balanced chemical equation, the number of all the atoms or elements exists on the reactant side equal to the number of all the atoms or elements present on the product side. Every measurement is dependent on the balanced equation. This is further used in determining the number of moles of reacting species, which is the key step in every calculation.

Answer to Problem 70E

The balanced form of chemical equation is $2{\text{MoS}}_2\left(s\right)+7{\text{O}}_2\left(g\right)\to 2{\text{MoO}}_{\text{3}}\left(s\right)+4{\text{SO}}_2\left(g\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{ MoS}}_2\left(s\right)+{\text{O}}_2\left(g\right)\to {\text{MoO}}_{\text{3}}\left(s\right)+{\text{SO}}_2\left(g\right)$

The number of oxygen atom on reactant side is two whereas on product side is five.

The number of sulfur atom on reactant side is two whereas on product side is one.

The number of molybdenum atom on reactant side and product side is one.

To balance the equation, add coefficient $2$ in front of ${\text{MoS}}_2\left(s\right)$ and $2$ in front of ${\text{MoO}}_{\text{3}}\left(s\right)$ .

The equation would become as follows.

  ${\text{ 2MoS}}_2\left(s\right)+{\text{O}}_2\left(g\right)\to 2{\text{MoO}}_{\text{3}}\left(s\right)+{\text{SO}}_2\left(g\right)$

The number of oxygen atom on reactant side is two and product side is eight.

The number of sulfur atom on reactant side is four whereas on product side is one.

The number of molybdenum atom on reactant side and product side is two.

To balance the equation, add coefficient $4$ in front of ${\text{SO}}_2\left(g\right)$ and $7$ in front of ${\text{O}}_2\left(g\right)$ .The equation would become as follows.

  ${\text{ 2MoS}}_2\left(s\right)+7{\text{O}}_2\left(g\right)\to 2{\text{MoO}}_{\text{3}}\left(s\right)+4{\text{SO}}_2\left(g\right)$

Thus, all the atoms of molybdenum, sulfur and oxygen on reactant and product side are balanced.

Therefore, the balanced chemical equation is. ${\text{ 2MoS}}_2\left(s\right)+7{\text{O}}_2\left(g\right)\to 2{\text{MoO}}_{\text{3}}\left(s\right)+4{\text{SO}}_2\left(g\right)$ .

(g)

Interpretation Introduction

Interpretation: The balanced form of chemical equation ${\text{ FeCO}}_3\left(s\right)+{\text{H}}_2{\text{CO}}_3\left(aq\right)\to \text{Fe}{\left({\text{HCO}}_3\right)}_2\left(aq\right)$ is to be stated.

Concept introduction: In any balanced chemical equation, the number of all the atoms or elements exists on the reactant side equal to the number of all the atoms or elements present on the product side. Every measurement is dependent on the balanced equation. This is further used in determining the number of moles of reacting species, which is the key step in every calculation.

Answer to Problem 70E

The balanced form of chemical equation is ${\text{ FeCO}}_3\left(s\right)+{\text{H}}_2{\text{CO}}_3\left(aq\right)\to \text{Fe}{\left({\text{HCO}}_3\right)}_2\left(aq\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{ FeCO}}_3\left(s\right)+{\text{H}}_2{\text{CO}}_3\left(aq\right)\to \text{Fe}{\left({\text{HCO}}_3\right)}_2\left(aq\right)$

The number of oxygen atom on reactant and product side is six.

The number of hydrogen atom on reactant and product side is two.

The number of iron atom on reactant and product side is one.

The number of carbon atom on reactant on reactant and product side is two.

Thus, the number of $\text{Fe,}\text{C}$ and $\text{O}$ atoms from reactant and product side are already same.

Therefore, the balanced chemical equation is ${\text{ FeCO}}_3\left(s\right)+{\text{H}}_2{\text{CO}}_3\left(aq\right)\to \text{Fe}{\left({\text{HCO}}_3\right)}_2\left(aq\right)$ .