Chapter 3, Problem 102E

Balance each of the following chemical equations.

a. KO₂(s) + H₂O(l) ⟶ KOH(aq) + O₂(g) + H₂O₂(aq)

b. Fe₂O₃(s) + HNO₃(aq) ⟶ Fe(NO₃)(aq) + H₂O(l)

c. NH₃(g) + O₂(g) ⟶ NO(g) + H₂O (g)

d. PCl₅(l) + H₂O(l) ⟶ H₃PO₄(aq) + HCl(g)

e. CaO(s) + C(s) ⟶ CaC₂(s) + CO₂ (g)

f. MoS₂(s) + O₂(g) ⟶ MoO₃(s) + SO₂ (g)

g. FeCO₃(s) + H₂CO₃(aq) ⟶ Fe(HCO₃)₂(aq)

Interpretation Introduction

Interpretation: A balanced form of the given chemical equations is to be stated.

Concept introduction: According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed during a chemical process.

To determine: A balanced form of the given chemical equation.

Explanation of Solution

According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed. Therefore a chemical equation, having lesser number of moles of an element on either side of a reaction, is balanced using appropriate numerical coefficients to satisfy the law of conservation of mass.

The given reaction is,

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

Adding the coefficient $2$ to ${\text{H}}_2\text{O}$ balances the number of atoms of hydrogen present on either side of the reaction. The equation thus obtained is,

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

Adding the coefficient $2$ to ${\text{KO}}_2$ and $\text{KOH}$ balances the number of atoms of potassium and oxygen present on either side of the reaction. The balanced equation thus obtained is,

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

Conclusion

The balanced chemical equation is,

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

Interpretation Introduction

Interpretation: A balanced form of the given chemical equations is to be stated.

Concept introduction: According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed during a chemical process.

To determine: A balanced form of the given chemical equation.

Explanation of Solution

According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed. Therefore a chemical equation, having lesser number of moles of an element on either side of a reaction, is balanced using appropriate numerical coefficients to satisfy the law of conservation of mass.

The given reaction is,

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

Adding the coefficient $2$ to $\text{Fe}{\left({\text{NO}}_3\right)}_3$ balances the number of atoms of iron present on either side of the reaction. The equation thus obtained is,

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

Adding the coefficient $6$ to ${\text{HNO}}_3$ and the coefficient $3$ to ${\text{H}}_2\text{O}$ balances the number of atoms of nitrogen, hydrogen and oxygen present on either side of the reaction. The balanced equation thus obtained is,

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

Conclusion

The balanced chemical equation is,

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

Interpretation Introduction

Interpretation: A balanced form of the given chemical equations is to be stated.

Concept introduction: According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed during a chemical process.

To determine: A balanced form of the given chemical equation.

Explanation of Solution

According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed. Therefore a chemical equation, having lesser number of moles of an element on either side of a reaction, is balanced using appropriate numerical coefficients to satisfy the law of conservation of mass.

The given reaction is,

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

Adding the coefficient $4$ to ${\text{NH}}_3$ and the coefficient $6$ to ${\text{H}}_2\text{O}$ balances the number of atoms of hydrogen present on either side of the reaction. The equation thus obtained is,

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

Adding the coefficient $5$ to ${\text{O}}_2$ and the coefficient $4$ to $\text{NO}$ balances the number of atoms of oxygen and nitrogen present on either side of the reaction. The balanced equation thus obtained is,

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

Conclusion

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

Interpretation Introduction

Interpretation: A balanced form of the given chemical equations is to be stated.

Concept introduction: According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed during a chemical process.

To determine: A balanced form of the given chemical equation.

Explanation of Solution

According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed. Therefore a chemical equation, having lesser number of moles of an element on either side of a reaction, is balanced using appropriate numerical coefficients to satisfy the law of conservation of mass.

The given reaction is,

${\text{PCl}}_5\left(l\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_3{\text{PO}}_4\left(aq\right)+\text{HCl}\left(g\right)$

Adding the coefficient $5$ to $\text{HCl}$ balances the number of atoms of chlorine present on either side of the reaction. The equation thus obtained is,

${\text{PCl}}_5\left(l\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_3{\text{PO}}_4\left(aq\right)+5\text{HCl}\left(g\right)$

Adding the coefficient $4$ to ${\text{H}}_{\text{2}}\text{O}$ balances the number of atoms of oxygen and hydrogen present on either side of the reaction. The balanced equation thus obtained is,

${\text{PCl}}_5\left(l\right)+4{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_3{\text{PO}}_4\left(aq\right)+5\text{HCl}\left(g\right)$

Conclusion

The balanced chemical equation is ${\text{PCl}}_5\left(l\right)+4{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_3{\text{PO}}_4\left(aq\right)+5\text{HCl}\left(g\right)$.

Interpretation Introduction

Interpretation: A balanced form of the given chemical equations is to be stated.

Concept introduction: According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed during a chemical process.

To determine: A balanced form of the given chemical equation.

Explanation of Solution

According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed. Therefore a chemical equation, having lesser number of moles of an element on either side of a reaction, is balanced using appropriate numerical coefficients to satisfy the law of conservation of mass.

The given reaction is,

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

Adding the coefficient $5$ to $\text{C}$ and the coefficient $2$ to ${\text{CaC}}_2$ balances the number of atoms of carbon present on either side of the reaction. The equation thus obtained is,

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

Adding the coefficient $2$ to $\text{CaO}$ balances the number of atoms of oxygen and calcium present on either side of the reaction. The balanced equation thus obtained is,

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

Conclusion

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

Interpretation Introduction

Interpretation: A balanced form of the given chemical equations is to be stated.

Concept introduction: According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed during a chemical process.

To determine: A balanced form of the given chemical equation.

Explanation of Solution

According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed. Therefore a chemical equation, having lesser number of moles of an element on either side of a reaction, is balanced using appropriate numerical coefficients to satisfy the law of conservation of mass.

The given reaction is,

${\text{MoS}}_2\left(s\right)+{\text{O}}_2\left(s\right)\to {\text{MoO}}_3\left(s\right)+{\text{SO}}_2\left(g\right)$

Adding the coefficient $2$ to ${\text{MoS}}_2$ and the coefficient $4$ to ${\text{SO}}_2$ balances the number of atoms of sulfur present on either side of the reaction. The equation thus obtained is,

$2{\text{MoS}}_2\left(s\right)+{\text{O}}_2\left(s\right)\to {\text{MoO}}_3\left(s\right)+4{\text{SO}}_2\left(g\right)$

Adding the coefficient $7$ to ${\text{O}}_2$ and the coefficient $4$ to ${\text{SO}}_2$ balances the number of atoms of oxygen and molybdenum present on either side of the reaction. The balanced equation thus obtained is,

$2{\text{MoS}}_2\left(s\right)+7{\text{O}}_2\left(s\right)\to 2{\text{MoO}}_3\left(s\right)+4{\text{SO}}_2\left(g\right)$

Conclusion

The balanced chemical equation is $2{\text{MoS}}_2\left(s\right)+7{\text{O}}_2\left(s\right)\to 2{\text{MoO}}_3\left(s\right)+4{\text{SO}}_2\left(g\right)$.

Interpretation Introduction

Interpretation: A balanced form of the given chemical equations is to be stated.

Concept introduction: According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed during a chemical process.

To determine: A balanced form of the given chemical equation.

Explanation of Solution

According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed. Therefore a chemical equation, having lesser number of moles of an element on either side of a reaction, is balanced using appropriate numerical coefficients to satisfy the law of conservation of mass.

The given reaction is,

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

The given reaction is already present in its balanced form.

Conclusion

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