Chapter 10, Problem 97QRT

Assume that a car burns pure octane.

C₈H₁₈ (d = 0.703 g/cm³)

(a) Write the balanced equation for burning octane in air. forming CO₂ and H₂O.

(b) If the car has a fuel efficiency of 32 miles per gallon of octane, what volume of CO₂ at 25 °C and 1.0 atm is generated when the car goes on a 10.0-mile trip? (The volume of 1 mol CO₂(g) at 25 °C and 1 atm is 24.5 L.)

Interpretation Introduction

Interpretation:

A balanced chemical equation for burning of octane has to be stated.

Concept Introduction:

The performance of an engine or fuel is measured in terms of octane number or octane rating.  The processes that are used to increase the octane number of a fuel are known as catalytic cracking and catalytic reforming.  Catalytic reforming converts the straight chain hydrocarbon into the branched hydrocarbon.

Answer to Problem 97QRT

A balanced chemical equation for burning of octane is shown below.

    ${\text{2C}}_8{\text{H}}_{18}\left(l\right)+25{\text{O}}_{\text{2}}\left(\text{g}\right)\to 16{\text{CO}}_2\left(\text{g}\right)+18{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Explanation of Solution

The chemical reaction for combustion of octane is as follows.

    ${\text{C}}_8{\text{H}}_{18}\left(l\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{CO}}_2\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

In the above equation, the number of atoms on reactant and product side is not balanced.  In order to balance them, a coefficient of $2$ is added with ${\text{C}}_8{\text{H}}_{18}$, a coefficient of $25$ is added with ${\text{O}}_{\text{2}}$, a coefficient of $16$ is added with ${\text{CO}}_2$ and a coefficient of $18$ is added with ${\text{H}}_{\text{2}}\text{O}$.

Therefore, the balanced chemical reaction for combustion of octane is shown below.

    ${\text{2C}}_8{\text{H}}_{18}\left(l\right)+25{\text{O}}_{\text{2}}\left(\text{g}\right)\to 16{\text{CO}}_2\left(\text{g}\right)+18{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Interpretation Introduction

Interpretation:

The volume of ${\text{CO}}_2$ obtained on a $\text{10}\text{mile}$ trip by car has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 97QRT

The volume of ${\text{CO}}_2$ obtained on a $\text{10}\text{mile}$ trip by car is $\underset{\_}{1426.88L}$.

Explanation of Solution

The amount of fuel that is octane required to travel $\text{32}\text{mile}$ by car is $\text{1}\text{gallon}$.  So, the amount of fuel required to travel $\text{10}\text{mile}$ by car is calculated as shown below.

    $\begin{array}{c}\text{32}\text{mile}=\text{1}\text{gallon}\\ \text{1}\text{mile}=\frac{\text{1}\text{gallon}}{\text{32}\text{mile}}\\ \text{10}\text{mile}=\frac{\text{1}\text{gallon}}{\text{32}\text{mile}}\times \text{10}\text{mile}\\ =\text{0}\text{.3125}\text{gallon}\end{array}$

The relation between gallon and liters is as follows.

    $\text{1}\text{gallon}=\text{3}\text{.7854}\text{L}$

Conversion of $\text{0}\text{.3125}\text{gallon}$ into $\text{L}$ is done as shown below.

    $\begin{array}{c}\text{1}\text{gallon}=\text{3}\text{.7854}\text{L}\\ \text{0}\text{.3125}\text{gallon}=\frac{\text{0}\text{.3125}\text{gallon}}{\text{1}\text{gallon}}\times \text{3}\text{.7854}\text{L}\\ =1.1829\text{L}\end{array}$

The density of octane is $0.703\text{g}{\text{cm}}^{-3}$.  Since $\text{1}000{\text{cm}}^{\text{3}}=1\text{L}$, therefore, density is equal to $703\text{g}{\text{L}}^{-1}$.

The relation between volume and mass is shown below.

    $\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

Substitute the density and volume of octane in the above equation.

    $\begin{array}{c}703\text{g}{\text{L}}^{-1}=\frac{\text{Mass}}{1.1829\text{L}}\\ \text{Mass}=703\text{g}{\text{L}}^{-1}\times 1.1829\text{L}\\ =831.58\text{g}\end{array}$

The molar mass of octane is $114.23\text{g}/\text{mol}$.  The relation between moles and mass is shown below.

    $\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the mass and molar mass of octane in the above equation.

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{831.58\text{g}}{114.23\text{g}/\text{mol}}\\ =7.28\text{mol}\end{array}$

From the balanced chemical equation for combustion of octane, it is known that one mole of octane gives $\text{8}\text{mol}$ of ${\text{CO}}_2$.

Therefore, the number of moles of ${\text{CO}}_2$ produced by $7.28\text{mol}$ of octane is calculated as shown below.

    $\begin{array}{c}\text{1}\text{mol}\text{of}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}=\text{8}\text{mol}\text{of}{\text{CO}}_{\text{2}}\\ 7.28\text{mol}\text{of}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}=\frac{\text{8}\text{mol}}{1\text{mol}}\times 7.28\text{mol}\text{of}{\text{CO}}_{\text{2}}\\ =58.24\text{mol}\text{of}{\text{CO}}_{\text{2}}\end{array}$

The volume of $\text{1}\text{mol}$ of ${\text{CO}}_2$ at $25°\text{C}$ and $\text{1}\text{atm}$ is $\text{24}\text{.5}\text{L}$.  Therefore, the volume of $58.24\text{mol}$ of ${\text{CO}}_2$ at $25°\text{C}$ and $\text{1}\text{atm}$ is $\left(\text{24}\text{.5}\times 58.24\right)\text{L}$ that is $\underset{\_}{1426.88L}$.