Question

(a) If an automobile travels 225 mi with a gas mileage of 20.5 mi/gal, how many kilograms of \(\mathrm{CO}_{2}\) are produced? Assume that the gasoline is composed of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l),\) whose density is 0.692 \(\mathrm{g} / \mathrm{mL}\) . (b) Repeat the calculation for a truck that has a gas mileage of 5 \(\mathrm{mi} / \mathrm{gal} .\)

Short answer

If an automobile travels 225 mi with a gas mileage of 20.5 mi/gal, it produces 96.62 kg of CO2, and if a truck travels the same distance with a gas mileage of 5 mi/gal, it produces 395.71 kg of CO2.

Step-by-step solution

Calculate Gallons of Octane Consumed

First, let's convert the miles into gallons of fuel consumed for both automobile and truck. For the automobile: 225 mi / 20.5 mi/gal = \(10.976\) gallons of octane consumed For the truck: 225 mi / 5 mi/gal = \(45\) gallons of octane consumed

Calculate Moles of Octane Consumed

Now let's convert the gallons of octane consumed into moles. First, we need to convert gallons to liters and then to milliliters (1 gal = 3.785 L and 1 L = 1000 mL). Now the volume can be converted to mass based on the given density (0.692 g/mL): The molar mass of octane C8H18 is: \(8 * 12.01 + 18 * 1.008 = 114.23\) g/mol For the automobile: \((10.976 \text{ gal})\left(\frac{3.785\text{L}}{1\text{ gal}}\right)\left(\frac{1000\text{mL}}{1\text{ L}}\right)\left(\frac{0.692 \text{g}}{1\text{ ml}}\right)\left(\frac{1 \text{ mol of octane}}{114.23\text{g}}\right) = 274.36 \text{ mol of octane} \) For the truck: \((45 \text{ gal})\left(\frac{3.785\text{L}}{1\text{ gal}}\right)\left(\frac{1000\text{mL}}{1\text{ L}}\right)\left(\frac{0.692 \text{g}}{1\text{ ml}}\right)\left(\frac{1 \text{ mol of octane}}{114.23\text{g}}\right) = 1,123.84 \text{ mol of octane} \)

Calculate Moles of CO2 Produced

Looking at the balanced chemical equation for the combustion of octane: \(2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O\) Moles of CO2 produced = 8 times moles of octane consumed For the automobile: \(274.36 \text{ mol of octane} * 8 = 2,195 \text{ mol of } CO_2\) For the truck: \(1,123.84 \text{ mol of octane} * 8 = 8,990.72 \text{ mol of } CO_2\)

Convert Moles of CO2 to Kilograms

Now, we will convert the moles of CO2 to the mass in kilograms. The molar mass of CO2 is: \(12.01 + 2 * 16.0 = 44.01\) g/mol For the automobile: \((2,195 \text{ mol } CO_2)\left(\frac{44.01\text{g}}{1\text{ mol }}\right)\left(\frac{1 \text{ kg }}{1000\text{g}}\right) = 96.62 \text{ kg of } CO_2\) For the truck: \((8,990.72 \text{ mol } CO_2)\left(\frac{44.01\text{g}}{1\text{ mol }}\right)\left(\frac{1 \text{ kg }}{1000\text{g}}\right) = 395.71 \text{ kg of } CO_2\) So, if an automobile travels 225 mi with a gas mileage of 20.5 mi/gal, it produces 96.62 kg of CO2, and if a truck travels the same distance with a gas mileage of 5 mi/gal, it produces 395.71 kg of CO2.

Key concepts

Combustion Reaction

A combustion reaction is a chemical process where a substance combines with oxygen to release energy. We often encounter it in daily life, especially when burning fuels like octane. During combustion, the fuel reacts with oxygen from the air, releasing heat and light.
Let's take the combustion of octane as an example. Octane, which is represented by the chemical formula \(C_8H_{18}\), is a component of gasoline. When octane burns, it reacts with oxygen to form carbon dioxide \(CO_2\) and water \(H_2O\). The equation of the octane combustion is:
\[2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O\]
This equation shows that two molecules of octane react with 25 molecules of oxygen to produce 16 molecules of carbon dioxide and 18 molecules of water. A lot of energy is released in the process, which our car engines and trucks use to move.

Octane

Octane is a hydrocarbon and is one of the main components in gasoline. It's known for its energy-rich nature because it has a lot of carbon and hydrogen atoms that can store energy within their chemical bonds. With the chemical formula \(C_8H_{18}\), octane reflects that each molecule consists of eight carbon atoms and eighteen hydrogen atoms.
When studied in terms of stoichiometry, octane's value to us can be calculated via its molar mass. The molar mass lets us go from the mass of octane used to the amount in moles. This step is crucial for balancing chemical equations and then predicting the products of combustion reactions.
Since octane is volatile, it combusts easily in the presence of oxygen, making it efficient for fuel. But, it's essential to balance its use due to its contribution to emissions. Knowing its density (e.g., \(0.692 \text{ g/mL}\) as given), we can also calculate the mass required for a given volume. This step is part of the process for estimating how much \(CO_2\) results from burning a certain amount of octane.

CO2 Emissions

Combustion reactions of hydrocarbons like octane lead to the release of carbon dioxide \(CO_2\), a significant greenhouse gas. Calculating \(CO_2\) emissions involves understanding the complete combustion process. Let's break it down:

• From the previous section about the combustion reaction, we've seen that the stoichiometry of the reaction is crucial.
• The balanced reaction equation tells us that each mole of octane results in eight moles of \(CO_2\) being emitted.
• Understanding the molar mass of \(CO_2\), which is \(44.01 \text{ g/mol}\), assists in converting the mole count to actual mass.

To measure \(CO_2\) emissions, we start by calculating the number of moles of octane burned. From there, using stoichiometric principles, we derive the moles of \(CO_2\) produced. Finally, we convert these moles to mass, often in kilograms, for practical and environmental assessments.
Reducing \(CO_2\) emissions from vehicles is a significant environmental priority. This can be achieved by using less fuel, opting for fuel-efficient vehicles, and considering alternative energy sources. Each step in a typical combustion equation links directly to calculating and understanding fuel consumption and its environmental impact.