Chapter 8, Problem 8.82QP

Interpretation Introduction

Interpretation: the efficiency of the given process should be determined when 1 gallon octane is burned and a total mass of 11.53kg of $\text{CO}$, ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ are produced.

Concept introduction:

• Balanced chemical equation of a reaction is written according to law of conservation of mass.
• Mole ratio between the reactant and a product of a reaction are depends upon the coefficients of reactant and product in a balanced chemical equation.
• Equation for Number of moles of a substance, from its given mass is,

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

• Number of grams of a substance from its number of moles is,

  $\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

Answer to Problem 8.82QP

The efficiency of the given process is $\text{86}\text{.53\%}$.

Explanation of Solution

In the given reactions,

Octane ( ${\text{C}}_8{\text{H}}_{18}$) undergoes complete as well incomplete combustion to produce ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ as well $\text{CO}$ and ${\text{H}}_2\text{O}$ respectively.

Therefore,

The chemical equations for these reactions are,

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

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

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Therefore,

The total number of each atom in the reactant side should equal to the total number of each atoms in the product side.

So, in order to balance a chemical equation, the coefficients of compounds or atoms are needed to be changed in such a way that total number of each atoms in the reactant side and the total number of each atoms in the product side is to become equal.

Hence,

The balanced equations for the given reactions are,

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

${\text{2C}}_8{\text{H}}_{18}\text{+}17{\text{O}}_2\to 16\text{CO}+18{\text{H}}_2\text{O}$

Let’s take mass of Octane ( ${\text{C}}_8{\text{H}}_{18}$) undergo complete combustion is x.

So, the number of moles of Octane ( ${\text{C}}_8{\text{H}}_{18}$) in this reaction is,

$\text{x g}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\text{×}\frac{\text{1mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}{\text{114}\text{.22g}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\text{=}\frac{\text{x}}{\text{114}\text{.22}}\text{mol}$.

The balanced chemical equation of the reaction is,

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

The mole ratio between ${\text{C}}_8{\text{H}}_{18}$ and ${\text{CO}}_{\text{2}}$ in the reaction is 1:8.

The mole ratio between ${\text{C}}_8{\text{H}}_{18}$ and ${\text{H}}_{\text{2}}\text{O}$ in the reaction is 1:9.

Therefore,

The number of moles of ${\text{CO}}_{\text{2}}$ produced in the reaction is $\frac{\text{8x}}{\text{114}\text{.22}}\text{mol}\text{=}\frac{\text{x}}{\text{14}\text{.2775}}\text{mol}$ and the number of moles of ${\text{H}}_{\text{2}}\text{O}$ produced in the reaction is $\frac{\text{9x}}{\text{114}\text{.22}}\text{mol}\text{=}\frac{\text{x}}{\text{12}\text{.6911}}\text{mol}$.

Then,

The mass of ${\text{CO}}_{\text{2}}$ is,

$\frac{\text{x}}{\text{14}\text{.2775}}\text{mol}{\text{CO}}_{\text{2}}\text{×}\text{44}\text{.01g}{\text{CO}}_{\text{2}}\text{=}\text{3}\text{.082x}\text{g}{\text{CO}}_{\text{2}}$.

The mass of ${\text{H}}_{\text{2}}\text{O}$ is,

$\frac{\text{x}}{\text{12}\text{.6911}}\text{mol}{\text{H}}_{\text{2}}\text{O}\text{×}\text{18}\text{.02g}{\text{H}}_{\text{2}}\text{O}\text{=}\text{1}\text{.42x}\text{g}{\text{H}}_{\text{2}}\text{O}$.

The total mass of octane undergoes combustion is given as $\text{1 gal}\text{×}\frac{\text{2}\text{.650kg}}{\text{1}\text{gal}}\text{=}\text{2650g}$.

Let’s take mass of Octane ( ${\text{C}}_8{\text{H}}_{18}$) undergo incomplete combustion is ‘2650-x’.

So, the number of moles of Octane ( ${\text{C}}_8{\text{H}}_{18}$) in this reaction is,

$\text{(2650-x) g}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\text{×}\frac{\text{1mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}{\text{114}\text{.22g}{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\text{=}\frac{\text{(2650-x)}}{\text{114}\text{.22}}\text{mol}$.

The balanced chemical equation of the reaction is,

${\text{2C}}_8{\text{H}}_{18}\text{+}17{\text{O}}_2\to 16\text{CO}+18{\text{H}}_2\text{O}$

The mole ratio between ${\text{C}}_8{\text{H}}_{18}$ and $\text{CO}$ in the reaction is 1:8.

The mole ratio between ${\text{C}}_8{\text{H}}_{18}$ and ${\text{H}}_{\text{2}}\text{O}$ in the reaction is 1:9.

Therefore,

The number of moles of $\text{CO}$ produced in the reaction is $\frac{\text{8(2650-x)}}{\text{114}\text{.22}}\text{mol}\text{=}\frac{\text{(2650-x)}}{\text{14}\text{.2775}}\text{mol}$.The number of moles of ${\text{H}}_{\text{2}}\text{O}$ produced in the reaction is $\frac{\text{9(2650-x)}}{\text{114}\text{.22}}\text{mol}\text{=}\frac{\text{(2650-x)}}{\text{12}\text{.6911}}\text{mol}$.

Then,

The mass of $\text{CO}$ is,

$\frac{\text{(2650-x)}}{\text{14}\text{.2775}}\text{mol}\text{CO}\text{×}\text{28}\text{.01g}\text{CO}\text{=}\text{(5199-1}\text{.962x)}\text{g}\text{CO}$.

The mass of ${\text{H}}_{\text{2}}\text{O}$ is,

$\frac{\text{(2650-x)}}{\text{12}\text{.6911}}\text{mol}{\text{H}}_{\text{2}}\text{O}\text{×}\text{18}\text{.02g}{\text{H}}_{\text{2}}\text{O}\text{=}\text{(3763-1}\text{.42x)}\text{g}{\text{H}}_{\text{2}}\text{O}$.

The following findings made by previous steps,

The mass of ${\text{CO}}_{\text{2}}$ is produced in the given complete combustion reaction is $\text{3}\text{.082x}\text{g}$.

The mass of ${\text{H}}_{\text{2}}\text{O}$ is produced in the given complete combustion reaction is $\text{1}\text{.42x}\text{g}$.

The mass of $\text{CO}$ is produced in the given incomplete combustion reaction is- $\text{(5199-1}\text{.962x)}\text{g}\text{CO}$.

The mass of ${\text{H}}_{\text{2}}\text{O}$ is produced in the given incomplete combustion reaction is- $\text{(3763-1}\text{.42x)}\text{g}{\text{H}}_{\text{2}}\text{O}$.

The total mass of $\text{CO}$, ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ produced by octane combustion is given that $\text{11}\text{.53kg}\text{=}\text{11530g}$.

Therefore,

$\begin{array}{l}\text{3}\text{.082x}\text{g}\text{+}\text{1}\text{.42x}\text{g}\text{+}\text{5199g}\text{-}\text{1}\text{.962x}\text{+}\text{3763g }\text{-}\text{ 1}\text{.42x}\text{=}\text{11530g}\\ \text{1}\text{.12x}\text{g}\text{+}\text{8962g}\text{=}\text{11530g}\\ \text{1}\text{.12x}\text{g}\text{=}\text{2568g}\\ \text{x}\text{=}\text{2293}\end{array}$

So, the mass of octane undergoes complete combustion or that converted to ${\text{CO}}_2$ is 2293g.

The total mass of octane undergoes combustion is given as $\text{2650g}$.

Hence,

The efficiency of the given process is,

$\frac{\text{2293g}}{\text{2650g}}\text{×}\text{100}\text{=}\text{86}\text{.53\%}$.

Conclusion

Conclusion

The efficiency of the given process of combustion of octane is determined according to the data’s given for the combustion reaction.