Chapter 3, Problem 3.155QP

Octane (C₈H₁₈) is a component of gasoline. Complete combustion of octane yields H₂O and CO₂. Incomplete combustion produces H₂O and CO, which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 1.000 gal of octane is burned in an engine. The total mass of CO, CO₂, and H₂O produced is 11.53 kg. Calculate the efficiency of the process; that is, calculate the fraction of octane converted to CO₂. The density of octane is 2.650 kg/gal.

Interpretation Introduction

Interpretation:

The efficiency of the given process should be determined when $1.000gallon$ octane is burned and a total mass of $11.53\text{kg}$$\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 3.155QP

  The efficiency of the given octane process is $\text{86}\text{.41\%}$.

Explanation of Solution

In the given chemical 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.

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,

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

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

The quantity of octane burned is $2650g$ (1 gallon with a density of $2.650\text{kg/gallon}$),

Let’$x$ be the mass of octane convert to ${\text{CO}}_{\text{2}}$ , therefore $(2650-x)g$ is the mass of octane converted to $(\text{CO})$

The amount of $\text{CO}$ and ${\text{H}}_{\text{2}}\text{O}$ produced by of octane are,

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 amount of $\text{CO}$ and ${\text{H}}_{\text{2}}\text{O}$ produced by $(2650-x)g$ of octane are,

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 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

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