Chapter 12, Problem 74A

Interpretation Introduction

Interpretation : The liters of air required to burn the gasoline are to be calculated.

Concept Introduction : The coefficients of a balanced chemical equation that are transformed into moles are used to calculate the mole ratio, which is a conversion factor.

When converting between a certain number of moles of one reactant or product and moles of another reactant or product, mole ratios are utilized.

The formula to calculate the number of moles is given:

  $\text{n}\text{=}\frac{\text{given}\text{mass}}{\text{molar mass}}$   ...... (1)

Answer to Problem 74A

The liters of air required to burn the gasoline is $1096493.33\text{L}$ .

Explanation of Solution

Given information:

A liter of gasoline goes $\text{9}\text{.2}\text{km}$ .

Gasoline is $\text{100\%}$ octane.

The density of octane is $\text{0}\text{.69}{\text{g/cm}}^3$ .

Air is $21\%$ oxygen.

Distance covered by gasoline is $1250\text{km}$ .

The amount of octane in $1250\text{km}$ trip is given as

  $\begin{array}{c}\text{V=}\frac{1250}{9.2}\\ =135.870\text{L}\\ \text{=135870}\text{mL}\end{array}$

The formula of mass is given as

  $\text{mass}\text{=}\text{density×volume}$

To calculate the mass of octane, substitute the values in the above formula:

  $\begin{array}{c}\text{mass}\text{=}\text{density×volume}\\ \text{= 0}\text{.69}{\text{g/cm}}^3\times 135870\text{mL}\left(\because 1\text{mL=1}{\text{cm}}^3\right)\\ =93750.3\text{g}\end{array}$

Molar mass of $\text{C}$ is $\text{12}\text{.00}\text{g/mol}$ .

Molar mass of $\text{H}$ is $1.00\text{g/mol}$ .

Mass of ${\text{C}}_8{\text{H}}_{18}$

(octane) is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{C}}_8{\text{H}}_{18}=8\text{M}\left(\text{C}\right)\text{+18M}\left(\text{H}\right)\\ =\left(\left(8\times 12.00\right)+\left(18\times 1.00\right)\right)\text{g/mol}\\ \text{= 114}\text{g/mol}\end{array}$

To calculate the moles of ${\text{C}}_8{\text{H}}_{18}$, substitute the values in equation (1):

  $\begin{array}{c}{\text{n}}_{{\text{C}}_8{\text{H}}_{18}}\text{=}\frac{{\text{m}}_{{\text{C}}_8{\text{H}}_{18}}}{{\text{M}}_{{\text{C}}_8{\text{H}}_{18}}}\\ =\frac{93750.3\cancel{\text{g}}}{114\cancel{\text{g}}\text{/mol}}\\ =822.37\text{mol}\end{array}$

The chemical equation involving octane with oxygen is given as follows:

  ${\text{2C}}_8{\text{H}}_{18}+25{\text{O}}_2\to 16{\text{CO}}_2+18{\text{H}}_2\text{O}$

So, $1\text{mole}$ of ${\text{C}}_8{\text{H}}_{18}$ contains $12.5\text{mole}$ of $1\text{mole}$ .

  $822.37\text{mol}$ of ${\text{C}}_8{\text{H}}_{18}$ is given as

  $822.37\text{mol}\times \text{12}\text{.5}\text{=}\text{10279}\text{.625}\text{mol}{\text{O}}_2$

Volume of $\text{1}\text{mole}$ of gas at STP contains $\text{22}\text{.4}\text{L}$ .

Volume of $\text{10279}\text{.625}\text{mol}{\text{O}}_2$ is given as

  $\text{10279}\text{.625}\text{mol}\times \text{22}\text{.4}\text{L}\text{=}\text{230263}\text{.6}\text{L}$

Air is $\text{21\%}$ oxygen, the volume of air given as

  $\begin{array}{c}\text{V=}\frac{230263.33\text{L}}{21\%}\\ =1096493.33\text{L}\end{array}$

The liters of air is $1096493.33\text{L}$ .