Chapter 4, Problem 4.94P

(a)

Interpretation Introduction

Interpretation:Ratio of ${\text{CO}}_{\text{2}}$ to $\text{CO}$ molecules in $5\text{ \%}$ of gasoline exhaust is to be calculated.

Concept introduction: Reactant when reacted in presence of oxygen molecule to form desired product such type of reaction is called combustion reaction. In this reaction, energy is produced in large amount with release of heat. Molecular oxygen is generally required for such kind of reaction that acts as an oxidizing agent.

The expression used for the calculation of number of moles of any species is as follows:

  $\text{Number of moles}=\frac{\text{mass}}{\text{molar mass}}$

Answer to Problem 4.94P

Ratio of ${\text{CO}}_{\text{2}}$ to $\text{CO}$ molecules in $5\text{ \%}$ of gasoline exhaust is $19$ .

Explanation of Solution

Incomplete combustion of gasoline, ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is given by the equation as follows:

  ${\text{2C}}_{\text{8}}{\text{H}}_{\text{18}}\left(l\right)+17{\text{O}}_{\text{2}}\left(g\right)\to 16\text{CO}\left(g\right)+18{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Complete combustion of gasoline, ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is given by the equation as follows:

  ${\text{2C}}_{\text{8}}{\text{H}}_{\text{18}}\left(l\right)+25{\text{O}}_{\text{2}}\left(g\right)\to 16{\text{CO}}_2\left(g\right)+18{\text{H}}_{\text{2}}\text{O}\left(g\right)$

  $5\text{ \%}$ of gasoline undergoes incomplete combustion to form $\text{CO}$ molecule. It means that out of total mass of gasoline that is $100\text{g}$ , $95\text{g}$ of gasoline undergoes complete combustion and $5\text{g}$ of gasoline undergoes incomplete combustion.

The expression used for the calculation of number of moles of gasoline is as follows:

  ${\text{ Moles of complete combustion C}}_{\text{8}}{\text{H}}_{\text{18}}=\frac{{\text{mass of complete combustion of C}}_{\text{8}}{\text{H}}_{\text{18}}}{{\text{molar mass of C}}_{\text{8}}{\text{H}}_{\text{18}}}$

Mass of complete combustion of ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $95\text{g}$ .

Molar mass of ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $114.22\text{ g/mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}{\text{ Moles of complete combustion C}}_{\text{8}}{\text{H}}_{\text{18}}=\frac{{\text{mass of complete combustion of C}}_{\text{8}}{\text{H}}_{\text{18}}}{{\text{molar mass of C}}_{\text{8}}{\text{H}}_{\text{18}}}\\ =\frac{95\text{g}}{114.22\text{ g/mol}}\\ =0.8317282\text{ mol}\end{array}$

The expression used for the calculation of number of moles of gasoline is as follows:

  ${\text{ Moles of incomplete combustion C}}_{\text{8}}{\text{H}}_{\text{18}}=\frac{{\text{mass of incomplete combustion of C}}_{\text{8}}{\text{H}}_{\text{18}}}{{\text{molar mass of C}}_{\text{8}}{\text{H}}_{\text{18}}}$

Mass of incomplete combustion of ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $5\text{g}$ .

Molar mass of ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $114.22\text{ g/mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}{\text{ Moles of incomplete combustion C}}_{\text{8}}{\text{H}}_{\text{18}}=\frac{{\text{mass of incomplete combustion of C}}_{\text{8}}{\text{H}}_{\text{18}}}{{\text{molar mass of C}}_{\text{8}}{\text{H}}_{\text{18}}}\\ =\frac{5\text{g}}{114.22\text{ g/mol}}\\ =0.0437751\text{ mol}\end{array}$

According to the balanced equation of complete combustion of gasoline, two moles of ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ reacts with oxygen to form sixteen moles of ${\text{CO}}_{\text{2}}$ . Therefore, the number of molecules of ${\text{CO}}_{\text{2}}$ is calculated as follows:

  ${\text{Molecules of CO}}_2=\left[\begin{array}{c}\left({\text{ Moles of complete combustion of C}}_{\text{8}}{\text{H}}_{\text{18}}\right)\\ \left(\frac{{\text{16 mol CO}}_2}{\text{2}{\text{mol C}}_{\text{8}}{\text{H}}_{\text{18}}}\right)\left(\frac{6.022\times {10}^{23}\text{molecules}}{1\text{mol}}\right)\end{array}\right]$

Moles of complete combustion of ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $0.8317282\text{ mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}{\text{Molecules of CO}}_2=\left[\begin{array}{c}\left({\text{ Moles of complete combustion C}}_{\text{8}}{\text{H}}_{\text{18}}\right)\\ \left(\frac{{\text{16 mol CO}}_2}{\text{2}{\text{mol C}}_{\text{8}}{\text{H}}_{\text{18}}}\right)\left(\frac{6.022\times {10}^{23}\text{molecules}}{1\text{mol}}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(0.8317282\text{ mol}\right)\left(\frac{{\text{16 mol CO}}_2}{\text{2}{\text{mol C}}_{\text{8}}{\text{H}}_{\text{18}}}\right)\\ \left(\frac{6.022\times {10}^{23}\text{molecules}}{1\text{mol}}\right)\end{array}\right]\\ =4.00693\times {10}^{24}\text{ molecules}\end{array}$

According to the balanced equation of incomplete combustion of gasoline, two moles of ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ reacts with oxygen to form sixteen moles of $\text{CO}$ . Therefore, number of molecules of $\text{CO}$ is calculated as follows:

  $\text{Molecules of CO}=\left[\begin{array}{c}\left({\text{ Moles of incomplete combustion of C}}_{\text{8}}{\text{H}}_{\text{18}}\right)\\ \left(\frac{\text{16 mol CO}}{\text{2}{\text{mol C}}_{\text{8}}{\text{H}}_{\text{18}}}\right)\left(\frac{6.022\times {10}^{23}\text{molecules}}{1\text{mol}}\right)\end{array}\right]$

Moles of incomplete combustion of ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $0.0437751\text{ mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Molecules of CO}=\left[\begin{array}{c}\left({\text{ Moles of incomplete combustion C}}_{\text{8}}{\text{H}}_{\text{18}}\right)\\ \left(\frac{{\text{16 mol CO}}_2}{\text{2}{\text{mol C}}_{\text{8}}{\text{H}}_{\text{18}}}\right)\left(\frac{6.022\times {10}^{23}\text{molecules}}{1\text{mol}}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(0.0437751\text{ mol}\right)\left(\frac{\text{16 mol}\text{CO}}{{\text{2 mol C}}_{\text{8}}{\text{H}}_{\text{18}}}\right)\\ \left(\frac{6.022\times {10}^{23}\text{molecules}}{1\text{mol}}\right)\end{array}\right]\\ =2.10891\times {10}^{23}\text{ molecules}\end{array}$

The ratio of molecules of ${\text{CO}}_{\text{2}}$ to $\text{CO}$ is as follows:

  $\text{Ratio}=\left(\frac{\text{Molecules of}{\text{CO}}_{\text{2}}}{\text{Molecules of}\text{CO}}\right)$

Molecules of ${\text{CO}}_{\text{2}}$ is $4.00693\times {10}^{24}\text{ molecules}$ .

Molecules of $\text{CO}$ is $2.10891\times {10}^{23}\text{ molecules}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Ratio}=\left(\frac{\text{Molecules of}{\text{CO}}_{\text{2}}}{\text{Molecules of}\text{CO}}\right)\\ =\left(\frac{4.00693\times {10}^{24}\text{ molecules}}{2.10891\times {10}^{23}\text{ molecules}}\right)\\ =18.99998\\ \approx 19\end{array}$

(b)

Interpretation Introduction

Interpretation:The ratio of masses of ${\text{CO}}_2$ to $\text{CO}$ is to be calculated.

Concept introduction:The quantity of species that gives the relation between reactants and products is determined by the stoichiometry of a reaction. Consider the general reaction,

  $\text{2A}+\text{B}\to 3\text{C}$

In the above reaction, two moles of $\text{A}$ reacts with one mole of $\text{B}$ to form three moles of $\text{C}$ . Therefore, stoichiometric ratio of A and B is $2:2$ , of A and C is $2:3$ and of B and C is $1:3$ .

Answer to Problem 4.94P

Ratio of mass of ${\text{CO}}_2$ to $\text{CO}$ is 30.

Explanation of Solution

The formula used for the calculation of mass of ${\text{CO}}_2$ is as follows:

  $\text{Mass of}{\text{CO}}_2=\left[\begin{array}{l}\text{moles of}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(\frac{16\text{mol}{\text{CO}}_2}{2\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\right)\\ \left({\text{molar mass of CO}}_2\right)\end{array}\right]$

Moles of ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $0.8317282\text{ mol}$ and molar mass of ${\text{CO}}_2$ is $44.01\text{ g/mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Mass of}{\text{CO}}_2=\left[\begin{array}{l}\text{moles of}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(\frac{16\text{mol}{\text{CO}}_2}{2\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\right)\\ \left({\text{molar mass of CO}}_2\right)\end{array}\right]\\ =\left[\left(0.8317282\text{ mol}\right)\left(\frac{16\text{mol}{\text{CO}}_2}{2\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\right)\left(44.01\text{ g/mol}\right)\right]\\ =292.83\text{g}\end{array}$

The formula used for the calculation of mass of $\text{CO}$ is as follows:

  $\text{Mass of}\text{CO}=\left[\begin{array}{l}\text{moles of}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(\frac{16\text{mol}\text{CO}}{2\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\right)\\ \left(\text{molar mass of CO}\right)\end{array}\right]$

Moles of ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $0.0437751\text{ mol}$ and molar mass of $\text{CO}$ is $28.01\text{ g/mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Mass of}\text{CO}=\left[\begin{array}{l}\text{moles of}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(\frac{16\text{mol}\text{CO}}{2\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\right)\\ \left(\text{molar mass of CO}\right)\end{array}\right]\\ =\left[\left(0.0437751\text{ mol}\right)\left(\frac{16\text{mol}\text{CO}}{2\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\right)\left(28.01\text{ g/mol}\right)\right]\\ =9.8091\text{g}\end{array}$

The expression used for the calculation of ratioof mass of ${\text{CO}}_{\text{2}}$ to $\text{CO}$ is as follows:

  $\text{Ratio}=\left(\frac{\text{Mass of}{\text{CO}}_{\text{2}}}{\text{Mass of}\text{CO}}\right)$

Mass of ${\text{CO}}_2$ is $292.83\text{g}$ and mass of $\text{CO}$ is $9.8091\text{g}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Ratio}=\left(\frac{\text{Mass of}{\text{CO}}_{\text{2}}}{\text{Mass of}\text{CO}}\right)\\ =\left(\frac{292.83\text{g}}{9.8091\text{g}}\right)\\ =29.85289\\ \approx 30\end{array}$

(c)

Interpretation Introduction

Interpretation:Percentage of gasoline that forms $\text{CO}$ for mass ratio of ${\text{CO}}_2$ to $\text{CO}$ to be exactly $\text{1/1}$ is to be calculated.

Concept introduction: Mass ratio of a component is defined as the ratio of mass of one component to that of other component. The ratio of two components A and B is expressed as follows:

  $\text{Mass ratio=}\left(\frac{\text{Mass of component A}}{\text{Mass of component B}}\right)$

Answer to Problem 4.94P

Percentage of gasoline that forms $\text{CO}$ for mass ratio of ${\text{CO}}_2$ to $\text{CO}$ to be exactly $\text{1/1}$ is $61\%$ .

Explanation of Solution

Suppose x be the fraction of ${\text{CO}}_2$ and fraction of $\text{CO}$ is $\text{1}-\text{x}$ .

The expression used for the calculation of mass ratio of ${\text{CO}}_2$ to $\text{CO}$ is as follows:

  $\text{Ratio}=\left(\frac{\left(\text{x}\right)\text{molar mass of}{\text{CO}}_{\text{2}}}{\left(\text{1}-\text{x}\right)\text{molar}\text{mass of}\text{CO}}\right)$

Molar mass of ${\text{CO}}_2$ is $44.01\text{ g/mol}$ , molar mass of $\text{CO}$ is $28.01\text{ g/mol}$ and ratio is 1.

Substitute the values in above equation.

  $\begin{array}{c}\text{Ratio}=\left(\frac{\left(\text{x}\right)\text{molar mass of}{\text{CO}}_{\text{2}}}{\left(\text{1}-\text{x}\right)\text{molar}\text{mass of}\text{CO}}\right)\\ \text{1}=\left(\frac{\left(\text{x}\right)\left(44.01\text{ g/mol}\right)}{\left(\text{1}-\text{x}\right)\left(28.01\text{ g/mol}\right)}\right)\\ 44.01\text{x}=28.01–28.01\text{x}\\ \text{x}=0.39\end{array}$

The value of x is $0.39$ and value of $\text{1}-\text{x}$ is $0.61$ . Therefore, percentage of gasoline that forms $\text{CO}$ for mass ratio of ${\text{CO}}_2$ to $\text{CO}$ to be exactly $\text{1/1}$ is $61\%$ .