Chapter 12, Problem 65P

Write a balanced equation for the combustion of each alkane
(a) ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{3}};\left(\text{b}\right){\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ .

Interpretation Introduction

(a)

Interpretation:

The balance equation for the combustion of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{3}}$ should be written.

Concept Introduction:

Compounds consist of carbon and hydrogen is known as hydrocarbons. Hydrocarbons are classified as saturated hydrocarbon and unsaturated hydrocarbon. Saturated hydrocarbons are those hydrocarbons in which carbon-carbon single bond is present as carbon is linked with four atoms. Unsaturated hydrocarbons are those hydrocarbons in which carbon-carbon multiple bonds are present that is double and triple bond.

Saturated hydrocarbon is known as alkane having general molecular formula ${\text{C}}_n{\text{H}}_{2n+2}$.

The chemical reaction which involves the reaction of alkane and oxygen, results in the formation of carbon dioxide gas and water vapor is known as combustion reaction. The combustion reaction is an exothermic reaction.

The general reaction of combustion of alkane is:

  ${\text{C}}_n{\text{H}}_{2n+2}+\left[n+\frac{(n+1)}{2}\right]{\text{O}}_{\text{2}}(g)\to n{\text{CO}}_{\text{2}}(g)+(n+1){\text{H}}_{\text{2}}\text{O}(g)$

A reaction is said to be balanced, if the number of atoms of each element on the product side and one the reactant side of a chemical reaction are equal.

Answer to Problem 65P

  ${\text{2CH}}_{\text{3}}{\text{CH}}_{\text{3}}(g)+7{\text{O}}_{\text{2}}(g)\to 4{\text{CO}}_{\text{2}}(g)+6{\text{H}}_{\text{2}}\text{O}(g)$

Explanation of Solution

Saturated hydrocarbon in which carbon-carbon single bond is present is known as alkane.

The general molecular formula of alkane is ${\text{C}}_n{\text{H}}_{2n+2}$ where n represents the number of carbon atoms.

The given alkane is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{3}}$ (ethane).

Ethane reacts with oxygen and results in the formation of carbon dioxide gas and water vapor.

The combustion reaction of ethane is written as:

  ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{3}}(g)+{\text{O}}_{\text{2}}(g)\to {\text{CO}}_{\text{2}}(g)+{\text{H}}_{\text{2}}\text{O}(g)$

Reactant side Product side

Number of C atoms = 2 Number of C atoms = 1

Number of H atoms = 6 Number of H atoms = 2

Number of O atoms = 2 Number of O atoms = 3

Thus, the above reaction is not balanced. To balance the reaction, multiply 2 with ethane, 7 with oxygen gas, 4 with carbon dioxide gas and 6 with water vapor.

Therefore, the balanced combustion reaction of ethane is written as:

  ${\text{2CH}}_{\text{3}}{\text{CH}}_{\text{3}}(g)+7{\text{O}}_{\text{2}}(g)\to 4{\text{CO}}_{\text{2}}(g)+6{\text{H}}_{\text{2}}\text{O}(g)$

Interpretation Introduction

(b)

Interpretation:

The balance equation for the combustion of ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ should be written.

Concept Introduction:

Compounds consist of carbon and hydrogen is known as hydrocarbons. Hydrocarbons are classified as saturated hydrocarbon and unsaturated hydrocarbon. Saturated hydrocarbons are those hydrocarbons in which carbon-carbon single bond is present as carbon is linked with four atoms. Unsaturated hydrocarbons are those hydrocarbons in which carbon-carbon multiple bonds are present that is double and triple bond.

Saturated hydrocarbon is known as alkane having general molecular formula ${\text{C}}_n{\text{H}}_{2n+2}$.

The chemical reaction which involves the reaction of alkane and oxygen, results in the formation of carbon dioxide gas and water vapor is known as combustion reaction. The combustion reaction is an exothermic reaction.

The general reaction of combustion of alkane is:

  ${\text{C}}_n{\text{H}}_{2n+2}+\left[n+\frac{(n+1)}{2}\right]{\text{O}}_{\text{2}}(g)\to n{\text{CO}}_{\text{2}}(g)+(n+1){\text{H}}_{\text{2}}\text{O}(g)$

A reaction is said to be balanced, if the number of atoms of each element on the product side and one the reactant side of a chemical reaction are equal.

Answer to Problem 65P

  ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}}(l)+8{\text{O}}_{\text{2}}(g)\to 5{\text{CO}}_{\text{2}}(g)+6{\text{H}}_{\text{2}}\text{O}(g)$

Explanation of Solution

Saturated hydrocarbon in which carbon-carbon single bond is present is known as alkane.

The general molecular formula of alkane is ${\text{C}}_n{\text{H}}_{2n+2}$ where n represents the number of carbon atoms.

The given alkane is ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ (2-methylbutane).

2-methylbutane reacts with oxygen and results in the formation of carbon dioxide gas and water vapor.

The combustion reaction of 2-methylbutane is written as:

  ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}}(l)+{\text{O}}_{\text{2}}(g)\to {\text{CO}}_{\text{2}}(g)+{\text{H}}_{\text{2}}\text{O}(g)$

Reactant side Product side

Number of C atoms = 5 Number of C atoms = 1

Number of H atoms = 12 Number of H atoms = 2

Number of O atoms = 2 Number of O atoms = 3

Thus, the above reaction is not balanced. To balance the reaction, multiply 8 with oxygen gas, 5 with carbon dioxide gas and 6 with water vapor.

Therefore, the balanced combustion reaction of2-methylbutaneis written as:

  ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}}(l)+8{\text{O}}_{\text{2}}(g)\to 5{\text{CO}}_{\text{2}}(g)+6{\text{H}}_{\text{2}}\text{O}(g)$