Question

Calculate the amount of carbon dioxide (in \(\mathrm{kg}\) ) emitted into the atmosphere by the complete combustion of a 15.0-gallon tank of gasoline. Do this by following these steps: a. Assume that gasoline is composed of octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\). Write a balanced equation for the combustion of octane. b. Determine the number of moles of octane contained in a 15.0-gallon tank of gasoline (1 gallon = \(3.78\) L). Octane has a density of \(0.79 \mathrm{~g} / \mathrm{mL}\). c. Use the balanced equation to convert from moles of octane to moles of carbon dioxide, then convert to grams of carbon dioxide, and finally to \(\mathrm{kg}\) of carbon dioxide.

Short answer

Using the balanced combustion equation, conversion factors, octane's density, and molar masses of octane and carbon dioxide, the amount of carbon dioxide emitted from a 15-gallon tank of gasoline is calculated in kilograms.

Step-by-step solution

Write Balanced Combustion Equation

To describe the combustion of octane, you write a chemical equation with octane \(\mathrm{C_8H_{18}}\) as the reactant, and carbon dioxide \(\mathrm{CO_2}\) and water \(\mathrm{H_2O}\) as the products, with oxygen \(\mathrm{O_2}\) as the required reactant for combustion. Then balance the equation. The balanced equation for the combustion of octane is \(2\mathrm{C_8H_{18}} + 25\mathrm{O_2} \rightarrow 16\mathrm{CO_2} + 18\mathrm{H_2O}\).

Calculate Moles of Octane

First convert 15.0 gallons of gasoline to liters using the conversion factor 1 gallon = 3.78 L, then to milliliters (1 L = 1000 mL). Then use the density of octane to convert milliliters to grams, and finally use the molar mass of octane \(\mathrm{C_8H_{18}}\) to convert grams to moles. The molar mass of octane \(\mathrm{C_8H_{18}}\) is \(114.23~\mathrm{g/mol}\).

Convert Moles of Octane to Moles of CO2

Using the stoichiometry of the balanced equation, convert the moles of octane to moles of carbon dioxide. According to the balanced equation, 2 moles of octane produce 16 moles of carbon dioxide. Use the ratio \(16 \mathrm{CO_2} / 2 \mathrm{C_8H_{18}}\) to find the moles of \(\mathrm{CO_2}\) produced.

Convert Moles of CO2 to Mass in Kilograms

First, convert the moles of carbon dioxide to grams using the molar mass of \(\mathrm{CO_2}\) (which is about \(44.01~\mathrm{g/mol}\)), and then convert grams to kilograms (1 kg = 1000 g).

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. When we apply stoichiometry to the combustion of octane, it allows us to predict the amount of products, like carbon dioxide and water, formed from a given amount of reactants. For instance, in the provided exercise, we use the stoichiometric coefficients from the balanced chemical equation to find the relationship between octane and carbon dioxide.

Let's imagine we are cooking with a recipe: stoichiometry tells us how much of each ingredient we need to make the dish, just like the balanced equation tells us how much carbon dioxide results from the combustion of a certain amount of octane. To solve the problem at hand, it's like figuring out how much cake we can bake with a given amount of flour. In this case, the octane is our 'flour,' and carbon dioxide is our 'cake.'

Chemical Equation Balancing

Balancing a chemical equation is essential to accurately convey the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. To balance an equation, we need to make sure that the number of atoms of each element is the same on both sides of the equation. It's like a seesaw: both sides need to be in equilibrium.

In the problem provided, we balance the combustion reaction of octane to ensure that we have equal numbers of each type of atom on the reactant side (octane and oxygen) and the product side (carbon dioxide and water). This balanced equation then serves as the blueprint for our stoichiometric calculations. Without a correctly balanced equation, our 'recipe' would be off, leading to either too much or too little of our 'cake' or carbon dioxide.

Mole-to-Mass Conversion

Converting between moles and mass is like exchanging currencies when traveling to different countries. Just as each currency has an exchange rate, each compound has a molar mass, which we use to convert between the 'currency' of moles and the 'currency' of grams. In our problem, once we know the number of moles of octane, we use its molar mass to convert it to grams. This step is crucial, as it bridges the gap between the abstract world of moles and the tangible world of mass.

Similarly, after we determine the moles of carbon dioxide produced, we convert them to grams using the molar mass of carbon dioxide, and then to kilograms. It's essential to get these conversions right to ensure accurate, real-world results – this is how we figure out the 'cost' of the emissions in kilograms of carbon dioxide for the combustion of gasoline in our engines.