Problem 72 - Chemistry: Principles and Reactions

Question

A Porsche \(928 \mathrm{~S} 4\) engine has a cylinder volume of \(618 \mathrm{~cm}^{3}\). The cylinder is full of air at \(75^{\circ} \mathrm{C}\) and \(1.00\) atm. (a) How many moles of oxygen are in the cylinder? (Mole percent of oxygen in air \(=21.0\).) (b) Assume that the hydrocarbons in gasoline have an average molar mass of \(1.0 \times 10^{2} \mathrm{~g} / \mathrm{mol}\) and react with oxygen in a \(1: 12 \mathrm{~mole}\) ratio. How many grams of gasoline should be injected into the cylinder to react with the oxygen?

Step-by-Step Solution

Answer

Answer: About 0.0378 grams of gasoline are required to react with the oxygen in the cylinder.

Step 1: Convert the given temperature to Kelvin

To use the Ideal Gas Law equation, we need to convert the given temperature from Celsius to Kelvin. We can do this using the following formula: \(K = C + 273.15\) where K is the temperature in Kelvin, and C is the temperature in Celsius. \(K = 75^{\circ} \mathrm{C} + 273.15 = 348.15 \mathrm{~K}\)

Step 2: Calculate the total number of moles of air in the cylinder

We can use the Ideal Gas Law equation, which states that: \(PV=nRT\) where P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the universal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin. Here, P = 1.00 atm, V = 0.618 L (convert 618 cm³ to L by dividing by 1000), and T = 348.15 K. Rearranging the Ideal Gas Law equation to solve for the number of moles (n): \(n=\frac{PV}{RT}\) Now, we can plug in the values and calculate the number of moles of air: \(n = \frac{(1.00 \mathrm{~atm})(0.618 \mathrm{~L})}{(0.0821 \mathrm{~L \cdot atm / mol \cdot K})(348.15 \mathrm{~K})} \approx 0.0216 \mathrm{~moles}\)

Step 3: Calculate the moles of oxygen in the cylinder

Since the mole percent of oxygen in the air is 21.0%, we can calculate the moles of oxygen in the cylinder as follows: Moles of oxygen = (Mole percent of oxygen / 100) × Total moles of air Moles of oxygen = (21.0 / 100) × 0.0216 = 0.004536 \mathrm{~moles}$

Step 4: Calculate the moles of gasoline required to react with the oxygen

According to the information given, the mole ratio of gasoline to oxygen is 1:12. This means for every 12 moles of oxygen, we need 1 mole of gasoline. Moles of gasoline = Moles of oxygen / 12 Moles of gasoline = 0.004536 / 12 = 0.000378 \mathrm{~moles}$

Step 5: Calculate the grams of gasoline required

Now that we have the moles of gasoline, we can calculate the grams required using the given average molar mass of hydrocarbons in gasoline (100 g/mol): Grams of gasoline = Moles of gasoline × Molar mass Grams of gasoline = 0.000378 \mathrm{~moles} × 1.0 × 10^{2} \mathrm{~g / mol} \approx 0.0378 \mathrm{~g}$ So, about 0.0378 grams of gasoline should be injected into the cylinder to react with the oxygen.

Key Concepts

Mole Calculation

Understanding how to calculate moles is crucial for grasping the relationship between mass, volume, and the number of particles in a sample of substance. A mole, often denoted by the unit 'mol', is a fundamental concept in chemistry that represents Avogadro's number \(6.02 \times 10^{23}\) of particles—atoms, molecules, ions, or electrons.

In the context of our exercise, mole calculation allows us to figure out how many moles of air and subsequently oxygen are in a Porsche \(928 \mathrm{~S} 4\) engine cylinder when given a volume, pressure, and temperature. Using the Ideal Gas Law \(PV=nRT\), we rearrange the formula to solve for \(n\), which stands for the number of moles. By inserting the pressure, volume in liters, the universal gas constant, and temperature in Kelvin, we can calculate the moles of air in the engine's cylinder.

For example, if the Porsche's engine cylinder is at \(75^\circ \mathrm{C}\) and \(1.00\) atm with a volume of \(618 \mathrm{~cm}^{3}\), converted to \(0.618 \mathrm{~L}\), the number of moles of air is calculated as \(n = \frac{{PV}}{{RT}}\). This foundational understanding of mole calculation is what allows us to delve deeper into the internal workings of engines and chemical reactions.

Oxygen in Air Composition

The air we breathe is a mixture of gases, with oxygen being one of the primary components essential for combustion and life. The composition of oxygen in air is approximately 21% by volume. This knowledge is not only interesting from a scientific perspective but also has practical applications, such as in calculating the amount of oxygen available for an engine to burn fuel.

In our exercise, this percentage allows us to use the total moles of air to find out the moles of oxygen present in the cylinder. By multiplying the total moles of air by the proportion of oxygen in air \(\text{Moles of oxygen} = \frac{21.0}{100} \times \text{Total moles of air}\), we get the moles of oxygen ready to react with gasoline during combustion. The accurate measurement of the oxygen percentage is vital in ensuring an efficient engine performance and is an application of stoichiometry, the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction.

Gasoline Combustion Reaction

Combustion reactions power most of our vehicles by burning fuel in the presence of oxygen to produce energy. In particular, the gasoline combustion reaction involves hydrocarbons reacting with oxygen to produce carbon dioxide, water, and energy. The balanced chemical equation for the combustion of gasoline (a hydrocarbon) can be represented as \(C_xH_y + O_2 \rightarrow CO_2 + H_2O + \text{energy}\). Understanding this reaction on a molecular level requires a clear grasp of mole ratios, as discussed in the exercise.

In the cylinder of the Porsche \(928 \mathrm{~S} 4\) engine, the hydrocarbons in gasoline react with oxygen in a specified mole ratio (in our case, 1:12). This ratio allows us to determine how much gasoline, in grams, needs to be injected into the cylinder. Using the calculated moles of oxygen and the average molar mass of gasoline provided, we can ensure the correct amount of fuel is used for efficient combustion and to minimize pollution.