Problem 28 - Chemistry The Central Science In Si Units

Question

In the contact process, sulfur dioxide and oxygen gas react to form sulfur trioxide as follows: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) $$ At a certain temperature and pressure, \(50 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) reacts with \(25 \mathrm{~L}\) of \(\mathrm{O}_{2}\). If all the \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are consumed, what volume of \(\mathrm{SO}_{3}\), at the same temperature and pressure, will be produced?

Step-by-Step Solution

Answer

50 liters of \(\mathrm{SO}_{3}\) will be produced.

Step 1: Identify the Reaction

Examine the chemical reaction given: \[2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\]. This indicates that 2 moles of \(\mathrm{SO}_{2}\) react with 1 mole of \(\mathrm{O}_{2}\) to produce 2 moles of \(\mathrm{SO}_{3}\).

Step 2: Determine the Limiting Reactant

According to the stoichiometry of the reaction, 2 volumes of \(\mathrm{SO}_{2}\) react with 1 volume of \(\mathrm{O}_{2}\). We have 50 L of \(\mathrm{SO}_{2}\) and 25 L of \(\mathrm{O}_{2}\), which perfectly match the stoichiometric ratio: \(\frac{50}{25} = 2\). Thus, there is no excess reactant.

Step 3: Apply Avogadro's Law

Avogadro's Law states that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules. Therefore, the volume ratios are equal to mole ratios in a gaseous reaction: \(2:1:2\) for \(\mathrm{SO}_{2}:\mathrm{O}_{2}:\mathrm{SO}_{3}\).

Step 4: Calculate Product Volume

Using the stoichiometry from the balanced equation, 50 L of \(\mathrm{SO}_{2}\) will produce 50 L of \(\mathrm{SO}_{3}\) since the ratio of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is 1:1.

Key Concepts

Limiting Reactant

In many chemical reactions, one reactant is entirely consumed before the others. This reactant is known as the "limiting reactant" because it limits the amount of product that can be formed. In our example, we have 50 liters of sulfur dioxide (SO₂) reacting with 25 liters of oxygen (O₂). According to the balanced equation, 2 volumes of SO₂ react with 1 volume of O₂:

• 2 parts SO₂ + 1 part O₂ → 2 parts SO₃

Given the quantities in the reaction, the SO₂ and O₂ are in perfect stoichiometric balance. That means neither reactant is in excess; all the SO₂ and O₂ are used entirely. In real-world reactions, finding the limiting reactant helps predict the amount of product formed, allowing chemists to optimize resources and reduce waste.

Avogadro's Law

Avogadro's Law plays a crucial role in solving problems involving gases. It states that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules. For stoichiometric calculations involving gases, this means we can use their volumes directly as their molar quantities. In our reaction:

• 2 parts SO₂ + 1 part O₂ → 2 parts SO₃

Since gases at the same conditions of temperature and pressure have the same number of molecules per volume, the coefficients in our balanced equation can be read as volumes. Thus, 50 L of SO₂ will require 25 L of O₂ to produce 50 L of SO₃. This simplification is only possible in reactions involving gases, demonstrating how Avogadro's Law facilitates straightforward stoichiometric calculations.

Gas Volume Ratios

Understanding gas volume ratios is key in reactions involving gases, especially under constant temperature and pressure. The balanced chemical equation provides insight into how the volumes of reactants relate to the volumes of products:

• 2 volumes of SO₂ + 1 volume of O₂ → 2 volumes of SO₃

In this exercise, the ratio of SO₂:O₂:SO₃ is 2:1:2. This means:

• 50 L of SO₂ will produce 50 L of SO₃, given 25 L of O₂ is completely consumed.

The gas volume ratios make the stoichiometry of gaseous reactions remarkably simple, as the volumes are directly proportional to the number of moles, thanks to Avogadro's Law. Always remember, these ratios and laws apply as long as the conditions of temperature and pressure remain consistent.