Chapter 10, Problem 10.97QP

(a)

Interpretation Introduction

Interpretation:

Balanced equation for the complete combustion of propane has to be written.

Concept Introduction:

Combustion reaction in general is said to be burning in presence of oxygen.  Alkanes undergo combustion to produce carbon dioxide and water along with a large amount of energy being released.  The general reaction scheme can be represented as,

  $\text{Alkane}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Explanation of Solution

Given hydrocarbon is propane.  Molecular formula of propane is ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$.  Alkanes and cycloalkanes on complete combustion produce carbon dioxide, water, and heat energy.  The chemical equation for the combustion of propane can be given as,

    ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing carbon atom:  In the product side, only one mol of carbon atom is present while in the reactant side, three mol of carbon atoms are present.  This can be balanced by adding coefficient 3 before ${\text{CO}}_{\text{2}}$ in the product side.  The chemical equation obtained after adding coefficient is,

    ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }3{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing hydrogen atoms:  In the product side, there are two mol of hydrogen atoms, while in the reactant side, there are eight mol hydrogen atoms.  Adding coefficient 4 before water in the product side balances out hydrogen atoms on both sides of equation.  The chemical equation obtained is,

    ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }3{\text{CO}}_{\text{2}}\text{+}4{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing oxygen atoms:  In the product side, there are ten mol oxygen atoms, while in the reactant side, there are only two mol oxygen atoms.  Adding coefficient 5 before ${\text{O}}_{\text{2}}$ in the reactant side balances out the oxygen atom on both sides of equation.  This step results in balanced chemical equation.  Therefore, the balanced chemical equation can be given as,

    ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{+}5{\text{O}}_{\text{2}}\stackrel{}{\to }3{\text{CO}}_{\text{2}}\text{+}4{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

(b)

Interpretation Introduction

Interpretation:

Balanced equation for the complete combustion of heptane has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given hydrocarbon is heptane.  Molecular formula of heptane is ${\text{C}}_7{\text{H}}_{\text{16}}$.  Alkanes and cycloalkanes on complete combustion produce carbon dioxide, water, and heat energy.  The chemical equation for the combustion of heptane can be given as,

    ${\text{C}}_7{\text{H}}_{\text{16}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing carbon atom:  In the product side, only one mol of carbon atom is present while in the reactant side, seven mol of carbon atoms are present.  This can be balanced by adding coefficient 7 before ${\text{CO}}_{\text{2}}$ in the product side.  The chemical equation obtained after adding coefficient is,

    ${\text{C}}_7{\text{H}}_{\text{16}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }7{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing hydrogen atoms:  In the product side, there are two mol of hydrogen atoms, while in the reactant side, there are sixteen mol hydrogen atoms.  Adding coefficient 8 before water in the product side balances out hydrogen atoms on both sides of equation.  The chemical equation obtained is,

    ${\text{C}}_7{\text{H}}_{\text{16}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }7{\text{CO}}_{\text{2}}\text{+}8{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing oxygen atoms:  In the product side, there are twenty two mol oxygen atoms, while in the reactant side, there are only two mol oxygen atoms.  Adding coefficient 11 before ${\text{O}}_{\text{2}}$ in the reactant side balances out the oxygen atom on both sides of equation.  This step results in balanced chemical equation.  Therefore, the balanced chemical equation can be given as,

    ${\text{C}}_7{\text{H}}_{\text{16}}\text{+}11{\text{O}}_{\text{2}}\stackrel{}{\to }7{\text{CO}}_{\text{2}}\text{+}8{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

(c)

Interpretation Introduction

Interpretation:

Balanced equation for the complete combustion of nonane has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given hydrocarbon is nonane.  Molecular formula of nonane is ${\text{C}}_9{\text{H}}_{\text{20}}$.  Alkanes and cycloalkanes on complete combustion produce carbon dioxide, water, and heat energy.  The chemical equation for the combustion of nonane can be given as,

    ${\text{C}}_9{\text{H}}_{\text{20}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing carbon atom:  In the product side, only one mol of carbon atom is present while in the reactant side, nine mol of carbon atoms are present.  This can be balanced by adding coefficient 9 before ${\text{CO}}_{\text{2}}$ in the product side.  The chemical equation obtained after adding coefficient is,

    ${\text{C}}_9{\text{H}}_{\text{20}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }9{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing hydrogen atoms:  In the product side, there are two mol of hydrogen atoms, while in the reactant side, there are twenty mol hydrogen atoms.  Adding coefficient 10 before water in the product side balances out hydrogen atoms on both sides of equation.  The chemical equation obtained is,

    ${\text{C}}_9{\text{H}}_{\text{20}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }9{\text{CO}}_{\text{2}}\text{+}10{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing oxygen atoms:  In the product side, there are twenty eight mol oxygen atoms, while in the reactant side, there are only two mol oxygen atoms.  Adding coefficient 14 before ${\text{O}}_{\text{2}}$ in the reactant side balances out the oxygen atom on both sides of equation.  This step results in balanced chemical equation.  Therefore, the balanced chemical equation can be given as,

    ${\text{C}}_9{\text{H}}_{\text{20}}\text{+}14{\text{O}}_{\text{2}}\stackrel{}{\to }9{\text{CO}}_{\text{2}}\text{+}10{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

(d)

Interpretation Introduction

Interpretation:

Balanced equation for the complete combustion of decane has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given hydrocarbon is decane.  Molecular formula of decane is ${\text{C}}_{10}{\text{H}}_{\text{22}}$.  Alkanes and cycloalkanes on complete combustion produce carbon dioxide, water, and heat energy.  The chemical equation for the combustion of decane can be given as,

    ${\text{C}}_{10}{\text{H}}_{\text{22}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing carbon atom:  In the product side, only one mol of carbon atom is present while in the reactant side, ten mol of carbon atoms are present.  This can be balanced by adding coefficient 10 before ${\text{CO}}_{\text{2}}$ in the product side.  The chemical equation obtained after adding coefficient is,

    ${\text{C}}_{10}{\text{H}}_{\text{22}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }10{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing hydrogen atoms:  In the product side, there are two mol of hydrogen atoms, while in the reactant side, there are twenty two mol hydrogen atoms.  Adding coefficient 11 before water in the product side balances out hydrogen atoms on both sides of equation.  The chemical equation obtained is,

    ${\text{C}}_{10}{\text{H}}_{\text{22}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }10{\text{CO}}_{\text{2}}\text{+}11{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing oxygen atoms:  In the product side, there are thirty one mol oxygen atoms, while in the reactant side, there are only two mol oxygen atoms.  Adding coefficient 15.5 before ${\text{O}}_{\text{2}}$ in the reactant side balances out the oxygen atom on both sides of equation.  This step results in balanced chemical equation.  Therefore, the balanced chemical equation can be given as,

    ${\text{C}}_{10}{\text{H}}_{\text{22}}\text{+}15.5{\text{O}}_{\text{2}}\stackrel{}{\to }10{\text{CO}}_{\text{2}}\text{+}11{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Multiplying the above equation by 2 gives whole numbers for all the compounds.  This can be given as,

    ${\text{2C}}_{10}{\text{H}}_{\text{22}}\text{+}31{\text{O}}_{\text{2}}\stackrel{}{\to }20{\text{CO}}_{\text{2}}\text{+}22{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$