Chapter 3, Problem 103E

Balance the following equations representing combustion reactions:

a.

b.

c. C₁₂H₂₂O₁₁ (s) + O₂(g) ⟶ CO₂ (g) + H₂O (g)

d. Fe(s) + O₂(g) ⟶ Fe₂O₃(s)

e. FeO(s) + O₂(g) ⟶ Fe₂O₃(s)

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,

${\text{C}}_6{\text{H}}_6\left(l\right)+{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

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

${\text{C}}_6{\text{H}}_6\left(l\right)+{\text{O}}_2\left(g\right)\to 6{\text{CO}}_2\left(g\right)+3{\text{H}}_2\text{O}\left(g\right)$

Adding the coefficient $\left(\frac{15}{2}\right)$ to ${\text{O}}_2$ balances the number of atoms of oxygen present on either side of the reaction. The equation thus obtained is,

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

Multiply the overall equation by $2$. The balanced chemical equation thus obtained is,

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

Conclusion

The balanced chemical equation is ${\text{2C}}_6{\text{H}}_6\left(l\right)+15{\text{O}}_2\left(g\right)\to 12{\text{CO}}_2\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)$.

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,

${\text{C}}_4{\text{H}}_{10}\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

Adding the coefficient $4$ to ${\text{CO}}_2$ and the coefficient $5$ to ${\text{H}}_2\text{O}$ balances the number of atoms of carbon and hydrogen present on either side of the reaction. The equation thus obtained is,

${\text{C}}_4{\text{H}}_{10}\left(g\right)+{\text{O}}_2\left(g\right)\to 4{\text{CO}}_2\left(g\right)+5{\text{H}}_2\text{O}\left(g\right)$

Adding the coefficient $\left(\frac{13}{2}\right)$ to ${\text{O}}_2$ balances the number of atoms of oxygen present on either side of the reaction. The equation thus obtained is,

${\text{C}}_4{\text{H}}_{10}\left(g\right)+\frac{13}{2}{\text{O}}_2\left(g\right)\to 4{\text{CO}}_2\left(g\right)+5{\text{H}}_2\text{O}\left(g\right)$

Multiply the overall equation by $2$. The balanced chemical equation thus obtained is,

${\text{2C}}_4{\text{H}}_{10}\left(g\right)+13{\text{O}}_2\left(g\right)\to 8{\text{CO}}_2\left(g\right)+10{\text{H}}_2\text{O}\left(g\right)$

Conclusion

The balanced chemical equation is ${\text{2C}}_4{\text{H}}_{10}\left(g\right)+13{\text{O}}_2\left(g\right)\to 8{\text{CO}}_2\left(g\right)+10{\text{H}}_2\text{O}\left(g\right)$.

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,

${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\left(s\right)+{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

Adding the coefficient $12$ to ${\text{CO}}_2$ and the coefficient $11$ to ${\text{H}}_2\text{O}$ balances the number of atoms of hydrogen and carbon present on either side of the reaction. The equation thus obtained is,

${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\left(s\right)+{\text{O}}_2\left(g\right)\to 12{\text{CO}}_2\left(g\right)+11{\text{H}}_2\text{O}\left(g\right)$

Adding the coefficient $12$ to ${\text{O}}_2$ balances the number of atoms of oxygen present on either side of the reaction. The balanced equation thus obtained is,

${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\left(s\right)+12{\text{O}}_2\left(g\right)\to 12{\text{CO}}_2\left(g\right)+11{\text{H}}_2\text{O}\left(g\right)$

Conclusion

The balanced chemical equation is ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\left(s\right)+12{\text{O}}_2\left(g\right)\to 12{\text{CO}}_2\left(g\right)+11{\text{H}}_2\text{O}\left(g\right)$.

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,

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

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

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

Adding the coefficient $\left(\frac{3}{2}\right)$ to ${\text{O}}_2$ balances the number of atoms of oxygen present on either side of the reaction. The equation thus obtained is,

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

Multiply the overall equation by $2$. The balanced chemical equation thus obtained is,

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

Conclusion

The balanced chemical equation is $\text{4Fe}\left(s\right)+3{\text{O}}_2\left(g\right)\to 2{\text{Fe}}_2{\text{O}}_3\left(s\right)$.

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,

$\text{FeO}\left(s\right)+{\text{O}}_2\left(g\right)\to {\text{Fe}}_2{\text{O}}_3\left(s\right)$

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

$\text{2FeO}\left(s\right)+{\text{O}}_2\left(g\right)\to {\text{Fe}}_2{\text{O}}_3\left(s\right)$

Adding the coefficient $\left(\frac{1}{2}\right)$ to ${\text{O}}_2$ balances the number of atoms of oxygen present on either side of the reaction. The equation thus obtained is,

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

Multiply the overall equation by $2$. The balanced chemical equation thus obtained is,

$\text{4FeO}\left(s\right)+{\text{O}}_2\left(g\right)\to 2{\text{Fe}}_2{\text{O}}_3\left(s\right)$

Conclusion

The balanced chemical equation is $\text{4FeO}\left(s\right)+{\text{O}}_2\left(g\right)\to 2{\text{Fe}}_2{\text{O}}_3\left(s\right)$