Chapter 10, Problem 98QRT

Interpretation Introduction

Interpretation:

The volume of ${\text{CO}}_2$ obtained on a $\text{10}\text{mile}$ trip by car using ${\text{CH}}_{\text{3}}\text{OH}$ has to be calculated.

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 98QRT

The volume of ${\text{CO}}_2$ obtained on a $\text{10}\text{mile}$ trip by car using ${\text{CH}}_{\text{3}}\text{OH}$ is $\underset{\_}{1144.88L}$.

Explanation of Solution

The amount of fuel that is ${\text{CH}}_{\text{3}}\text{OH}$ required to travel $\text{20}\text{mile}$ 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.

    $\begin{array}{c}\text{20}\text{mile}=\text{1}\text{gallon}\\ \text{1}\text{mile}=\frac{\text{1}\text{gallon}}{\text{20}\text{mile}}\\ \text{10}\text{mile}=\frac{\text{1}\text{gallon}}{20\text{mile}}\times \text{10}\text{mile}\\ =\text{0}\text{.5}\text{gallon}\end{array}$

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.

    $\begin{array}{c}\text{1}\text{gallon}=\text{3}\text{.7854}\text{L}\\ \text{0}\text{.5}\text{gallon}=\frac{\text{0}\text{.5}\text{gallon}}{\text{1}\text{gallon}}\times \text{3}\text{.7854}\text{L}\\ =1.8927\text{L}\end{array}$

The density of ${\text{CH}}_{\text{3}}\text{OH}$ is $0.791\text{g}{\text{cm}}^{-3}$.  Since $\text{1}000{\text{cm}}^{\text{3}}=1\text{L}$, therefore, density is equal to $791\text{g}{\text{L}}^{-1}$.

The relation between volume and mass is shown below.

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

Substitute the density and volume of ${\text{CH}}_{\text{3}}\text{OH}$ in the above equation.

    $\begin{array}{c}791\text{g}{\text{L}}^{-1}=\frac{\text{Mass}}{1.8927\text{L}}\\ \text{Mass}=791\text{g}{\text{L}}^{-1}\times 1.8927\text{L}\\ =1497.13\text{g}\end{array}$

The molar mass of ${\text{CH}}_{\text{3}}\text{OH}$ is $32.04\text{g}/\text{mol}$.  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 ${\text{CH}}_{\text{3}}\text{OH}$ in the above equation.

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{1497.13\text{g}}{32.04\text{g}/\text{mol}}\\ =46.73\text{mol}\end{array}$

The combustion of one mole of ${\text{CH}}_{\text{3}}\text{OH}$ gives one mole of ${\text{CO}}_2$.

Therefore, the number of moles of ${\text{CO}}_2$ produced by $46.73\text{mol}$ of ${\text{CH}}_{\text{3}}\text{OH}$ is equal to $46.73\text{mol}$.

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 $46.73\text{mol}$ of ${\text{CO}}_2$ at $25°\text{C}$ and $\text{1}\text{atm}$ is $\left(\text{24}\text{.5}\times 46.73\text{mol}\right)\text{L}$ that is $\underset{\_}{1144.88L}$.