Question

A cylinder fitted with a movable piston initially contains \(2.00 \mathrm{~mol}\) \(\mathrm{O}_{2}(g)\) and an unknown amount of \(\mathrm{SO}_{2}(g)\). The oxygen is known to be in excess. The density of the mixture is \(0.8000 \mathrm{~g} / \mathrm{L}\) at some \(T\) and \(P\). After the reaction has gone to completion, forming \(\mathrm{SO}_{3}(g)\), the density of the resulting gaseous mixture is \(0.8471 \mathrm{~g} / \mathrm{L}\) at the same \(T\) and \(P\). Calculate the mass of \(\mathrm{SO}_{3}\) formed in the reaction.

Short answer

The mass of SO3 formed in the reaction is 0 g. This is because there was no SO2 initially present in the mixture, so the reaction will not occur, and thus, no SO3 will be formed.

Step-by-step solution

Write the balanced chemical equation

First, we need to write down the balanced chemical equation for the reaction: O2(g) + 2 SO2(g) -> 2 SO3(g) Knowing the balanced chemical equation, we'll be able to relate moles and masses of reactants and products.

Calculate the moles of O2

We are given the amount of O2 in moles (2.00 moles).

Find the volume of the mixture

We are given the density of the initial mixture (0.8000 g/L) and total moles of O2. From this, we can calculate the volume of the mixture. First, let's find the mass of O2. Molar mass of O2 = 32.00 g/mol Mass of O2 = moles × molar mass = 2.00 mol × 32.00 g/mol = 64.00 g Now we can calculate the volume of the initial mixture using the formula: Volume = Mass / Density = 64.00 g / 0.8000 g/L = 80.00 L

Find the moles of SO2

We will now find the moles of SO2 present in the initial mixture. We can use the ideal gas law to find this because the same volume applies to both O2 and SO2 at the same temperature and pressure. With the volume and density given, we can find the total mass of the initial mixture: Total mass of the mixture = Volume × density = 80.00 L × 0.8000 g/L = 64.00 g Since we know the mass of O2 (64.00 g), we can find the mass of SO2: Mass of SO2 = Total mass of the mixture - Mass of O2 = 64.00 g - 64.00 g = 0 g In this case, there is no SO2 gas present in the initial mixture.

Determine the mass of SO3 formed

Since there was no SO2 initially, the reaction will not occur. As a result, there will be no SO3 formed in the reaction. Mass of SO3 formed = 0 g This is the final answer to the problem. The mass of SO3 formed in the reaction is 0 g.

Key concepts

Balanced Chemical Equation

A balanced chemical equation is essential in solving chemical reaction problems because it shows the relationship between reactants and products. For the reaction in this exercise, the chemical equation is given as:

• \( \text{O}_2(g) + 2 \text{SO}_2(g) \rightarrow 2 \text{SO}_3(g) \)

This equation tells us that one mole of oxygen reacts with two moles of sulfur dioxide to form two moles of sulfur trioxide. Balancing the chemical equation ensures that the number of atoms for each element is the same on both sides of the equation, adhering to the law of conservation of mass. This law states that atoms are neither lost nor created in chemical reactions. Understanding the balanced equation is crucial for calculating the relationship between reactants and products in terms of moles and masses.

Mole Calculation

Moles are a measure of the number of particles in a substance. In this exercise, we know there's 2.00 moles of \(\text{O}_2\). Moles help us use the stoichiometric coefficients in a balanced chemical equation to determine the proportions of reactants and products involved. With 2.00 moles of \(\text{O}_2\), we use these stoichiometry relationships to calculate the theoretical masses and volumes.To find the amount of a reactant or product, you multiply the number of moles by its molar mass. For example, the molar mass of \(\text{O}_2\) is 32.00 g/mol. To calculate mass, you use the formula:

• \( \text{Mass} = \text{Moles} \times \text{Molar Mass} \)

This formula gives essential input for calculating gas density and applying the ideal gas law in later steps.

Gas Density

Gas density is defined as the mass of gas per unit volume. It is an important physical property that helps determine the volume or mass of a gas in a particular setting. Given that the density of the gas mixture is 0.8000 g/L, we can calculate the volume if the mass is known. Using density in calculations:

• \(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\)
• \(\text{Volume} = \frac{\text{Mass}}{\text{Density}}\)

In chemical reactions, density can also inform about the changes in mass and volume before and after the reaction, aiding the understanding of reaction completion and composition adjustments. These calculations are critical, especially when combining gases as in the question above.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry linking temperature, pressure, volume, and number of moles of a gas. The equation is usually expressed as:

• \( PV = nRT \)

Where:

• \( P \) is the pressure in atm
• \( V \) is the volume in liters
• \( n \) is the number of moles
• \( R \) is the ideal gas constant (0.0821 L·atm/mol·K)
• \( T \) is the temperature in Kelvin

The ideal gas law can help us analyze the conditions of the gas mixture. It assumes the gases behave ideally, meaning no interactions occur between the molecules and volume taken by them is negligible compared to the container volume. In solving gas mixture problems, it’s crucial to apply this law to find unknowns such as volume or moles when the other properties are given. Although this textbook solution did not require a deep dive into these calculations due to zero initial SO₂, understanding the ideal gas law is essential in more involved chemical calculations.