Chapter 20, Problem 86A

(a)

Interpretation Introduction

Interpretation: The table is to be completed.

Concept Introduction: The only two types of atoms that make up all of an organic complex called a hydrocarbon are carbon and hydrogen. Gases called hydrocarbons are colourless and have hardly detectable odours.

Answer to Problem 86A

The completed table is as follows:

Alkane burned	${\text{O}}_{\text{2}}$ used (mol)	${\text{CO}}_{\text{2}}$ produced (mol)	${\text{H}}_{\text{2}}\text{O}$ produced (mol)
${\text{CH}}_{\text{4}}$	$\text{2}$	$\text{1}$	$\text{2}$
${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$	$3.5$	$\text{2}$	$\text{3}$
${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$	$5$	$\text{3}$	$4$
${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$	$6.5$	$4$	$5$
${\text{C}}_5{\text{H}}_{\text{12}}$	$8$	$5$	$6$
${\text{C}}_6{\text{H}}_{\text{14}}$	$9.5$	$6$	$7$

Explanation of Solution

Alkane when burnt with oxygen produces carbon dioxide and water.

From the hydrocarbons such as ${\text{CH}}_{\text{4}}$ , ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ and ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ , the relation between moles of carbon dioxide, hydrogen and oxygen is given as

  ${\text{O}}_{\text{2}}{\text{=CO}}_{\text{2}}\text{+}\frac{\text{1}}{\text{4}}\text{H}$

The moles of oxygen are equal to the sum of the carbon dioxide and one-fourth of the hydrogen.

The reaction of hydrocarbon ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$

with oxygen is given as follows:

  ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{+6}{\text{.5O}}_{\text{2}}\to {\text{4CO}}_{\text{2}}{\text{+5H}}_{\text{2}}\text{O}$

So, the value of $\text{a}$ is $\text{6}\text{.5}$ , $\text{b}$ is $\text{4}$ and $\text{c}$ is $\text{5}$ .

The reaction of hydrocarbon ${\text{C}}_5{\text{H}}_{\text{12}}$

with oxygen is given as follows:

  ${\text{C}}_5{\text{H}}_{\text{12}}{\text{+8O}}_{\text{2}}\to 5{\text{CO}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O}$

So, the value of $\text{d}$ is $\text{8}$ , $\text{e}$ is $\text{5}$ and $\text{f}$ is $6$ .

The reaction of hydrocarbon ${\text{C}}_6{\text{H}}_{\text{14}}$

with oxygen is given as follows:

  ${\text{C}}_6{\text{H}}_{\text{14}}\text{+9}{\text{.5O}}_{\text{2}}\to 6{\text{CO}}_{\text{2}}{\text{+7H}}_{\text{2}}\text{O}$

So, the value of $\text{g}$ is $9.5$ , $\text{h}$ is $6$ and $\text{i}$ is $7$ .

(b)

Interpretation Introduction

Interpretation: The balanced generalized equation for the complete oxidation of alkane is to be written. The coefficients in terms of x and y are to be written.

Concept Introduction: The only two types of atoms that make up all of an organic complex called a hydrocarbon are carbon and hydrogen. Gases called hydrocarbons are colourless and have hardly detectable odours.

Answer to Problem 86A

The generalized balanced chemical equation for the hydrocarbon combustion is given as follows:

  ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\text{+}\left(\text{x+}\left(\frac{\text{y}}{\text{4}}\right)\right){\text{O}}_{\text{2}}\to {\text{xCO}}_{\text{2}}\text{+}\left(\frac{\text{y}}{\text{2}}\right){\text{H}}_{\text{2}}\text{O}$

The coefficient of ${\text{O}}_{\text{2}}$ is $\text{x+}\frac{\text{y}}{\text{4}}$ .

The coefficient of ${\text{CO}}_{\text{2}}$ is $\text{x}$ .

The coefficient of ${\text{H}}_{\text{2}}\text{O}$ is $\frac{\text{y}}{2}$ .

Explanation of Solution

Alkane when burnt with oxygen produces carbon dioxide and water.

From the hydrocarbons such as ${\text{CH}}_{\text{4}}$ , ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ and ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ , the relation between moles of carbon dioxide, hydrogen and oxygen is given as

  ${\text{O}}_{\text{2}}{\text{=CO}}_{\text{2}}\text{+}\frac{\text{1}}{\text{4}}\text{H}$

The moles of oxygen are equal to the sum of the carbon dioxide and one-fourth of the hydrogen.

The generalized balanced chemical equation for the hydrocarbon combustion is given as follows:

  ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\text{+}\left(\text{x+}\left(\frac{\text{y}}{\text{4}}\right)\right){\text{O}}_{\text{2}}\to {\text{xCO}}_{\text{2}}\text{+}\left(\frac{\text{y}}{\text{2}}\right){\text{H}}_{\text{2}}\text{O}$

The coefficient of ${\text{O}}_{\text{2}}$ is $\text{x+}\frac{\text{y}}{\text{4}}$ .

The coefficient of ${\text{CO}}_{\text{2}}$ is $\text{x}$ .

The coefficient of ${\text{H}}_{\text{2}}\text{O}$ is $\frac{\text{y}}{2}$ .