Question

a. Write the balanced equation for the combustion of isooctane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) to produce water vapor and carbon dioxide gas. b. Assuming gasoline is \(100 . \%\) isooctane, with a density of 0.692 \(\mathrm{g} / \mathrm{mL}\) , what is the theoretical yield of carbon dioxide produced by the combustion of \(1.2 \times 10^{10}\) gal of gasoline (the approximate annual consumption of gasoline in the United States)?

Short answer

The balanced chemical equation for the combustion of isooctane to produce water vapor and carbon dioxide gas is: \(C_8H_{18}(l) + 12.5O_2(g) \rightarrow 8CO_2(g) + 9H_2O(g)\) The theoretical yield of carbon dioxide produced by the combustion of \(1.2 \times 10^{10}\ gal\) of gasoline (the approximate annual consumption of gasoline in the United States) is approximately \(5.87 \times 10^9 \ gal\).

Step-by-step solution

Write the unbalanced equation for the combustion of isooctane

The unbalanced chemical equation for the combustion of isooctane to form water vapor and carbon dioxide gas is: \(C_8H_{18}(l) + O_2(g) \rightarrow CO_2(g) + H_2O(g)\)

Balance the chemical equation

To balance the chemical equation, we will adjust the coefficients of the reactants and products accordingly: \(C_8H_{18}(l) + 12.5O_2(g) \rightarrow 8CO_2(g) + 9H_2O(g)\) Thus, the balanced chemical equation for the combustion of isooctane to produce water vapor and carbon dioxide gas is: \(C_8H_{18}(l) + 12.5O_2(g) \rightarrow 8CO_2(g) + 9H_2O(g)\) #b. Calculate the theoretical yield of carbon dioxide produced by the combustion of a given amount of gasoline#

Convert the volume of gasoline to grams

We are given the following information: - Gasoline volume = \(1.2 \times 10^{10}\) gal - Gasoline density = 0.692 g/mL - 1 gal = 3.78541 L = 3785.41 mL First, we convert the gasoline volume from gallons to milliliters: \(1.2 \times 10^{10}\ gal \times \frac{3.78541 \times 10^3 \ mL}{1 \ gal} = 4.542 \times 10^{13}\ mL\) Now, we convert the gasoline volume in milliliters to grams using the gasoline density: \(4.542 \times 10^{13}\ mL \times \frac{0.692 \ g}{1 \ mL} = 3.1414 \times 10^{13} \ g\)

Calculate the mass of CO₂ produced using stoichiometry

Based on the balanced chemical equation, 1 mol of \(C_8H_{18}\) produces 8 mol of \(CO_2\). We will use stoichiometry to calculate the mass of \(CO_2\) produced: Molar mass of \(C_8H_{18}\) = 8(12.01 g/mol C) + 18(1.008 g/mol H) = 114.23 g/mol Moles of \(C_8H_{18}\) = \(\frac{3.1414 \times 10^{13} \ g}{114.23 \ g/mol}\) = \(2.7482 \times 10^{11}\ mol\) Moles of \(CO_2\) produced = 8 mol \(CO_2\) × \(2.7482 \times 10^{11}\ mol \, C_8H_{18}\) = \(2.1986 \times 10^{12}\ mol\) Molar mass of \(CO_2\) = 12.01 g/mol C + 2(16.00 g/mol) = 44.01 g/mol Mass of \(CO_2\) produced = \(2.1986 \times 10^{12}\ mol \times \frac{44.01 \ g}{1 \ mol} = 9.6787 \times 10^{13} \ g\)

Express the mass of CO₂ produced in the required format

We are asked to find the theoretical yield of carbon dioxide in the form of "amount of gallons consumed" which is given in the problem. So, we need to convert the mass of \(CO_2\) back to the volume of gasoline: Mass of \(CO_2\) produced = 9.6787 × 10¹³ g Gasoline density = 0.692 g/mL Molar mass of \(C_8H_{18}\) = 114.23 g/mol Molar mass of \(CO_2\) = 44.01 g/mol So, 1 g of gasoline produces approximately \(8 \times \frac{1}{0.692} \times \frac{44.01}{114.23} \approx 4.353\) g of \(CO_2\). Now, we can find the volume of gasoline that produces the calculated mass of \(CO_2\): Volume of gasoline consumed = \(\frac{9.6787 \times 10^{13} \ g}{4.353 \ g/mL} = 2.2226 \times 10^{13} \ mL\) Finally, we convert the volume of gasoline to gallons: \(2.2226 \times 10^{13} \ mL \times \frac{1 \ gal}{3.78541 \times 10^3\ mL} = 5.8706 \times 10^{9} \ gal\) Thus, the theoretical yield of carbon dioxide produced by the combustion of \(1.2 \times 10^{10}\ gal\) of gasoline (the approximate annual consumption of gasoline in the United States) is approximately \(5.87 \times 10^9 \ gal\).

Key concepts

Isooctane

Isooctane, also known as 2,2,4-trimethylpentane, is a crucial component in gasoline. It's a hydrocarbon with the chemical formula \(\text{C}_8\text{H}_{18}\).
It is known for its ability to resist knocking in engines, which makes it a standard reference fuel for measuring octane ratings.
Octane ratings are used to indicate the fuel’s ability to resist pre-ignition or knocking during combustion. A higher octane number means better resistance to knocking.
In combustion reactions, isooctane efficiently combines with oxygen to release energy, making it an excellent fuel choice. Its structural properties contribute to its effectiveness in engines.
Understanding the role of isooctane in combustion is essential for analyzing fuel efficiency and performance in automotive applications.

Theoretical Yield

Theoretical yield is a crucial concept in chemistry that refers to the maximum amount of product that can be obtained from a chemical reaction, assuming complete efficiency with no losses. It serves as an ideal benchmark to measure the efficiency of a reaction.
To calculate theoretical yield, the stoichiometry of the balanced chemical equation is used. This involves determining the molar ratios of reactants and products.
When given the mass or volume of reactants, you convert these quantities into moles using their molar masses or densities.

• For example, when calculating the theoretical yield of carbon dioxide from the combustion of isooctane, the amount of isooctane used can be translated into moles.
• Then, using the balanced equation, you predict how many moles of carbon dioxide will be produced.

Understanding theoretical yield helps in evaluating the efficiency and economic viability of reactions in industrial processes.

Balanced Chemical Equation

A balanced chemical equation is essential for accurately describing a chemical reaction. The goal is to ensure that the number of atoms for each element is conserved on both sides of the equation.
Balancing involves adjusting the coefficients (the numbers in front of each compound or molecule) without changing the chemical identities.
For example, in the combustion reaction of isooctane \[\text{C}_8\text{H}_{18}(l) + 12.5 \text{O}_2(g) \rightarrow 8\text{CO}_2(g) + 9\text{H}_2\text{O}(g)\] This equation is balanced because there are equal numbers of carbon, hydrogen, and oxygen atoms on both sides.
Balancing ensures that matter is conserved according to the Law of Conservation of Mass.

• This is vital for making accurate predictions about the amounts of products formed and reactants consumed.

This knowledge is essential for performing stoichiometric calculations and for understanding the reaction's dynamics.