Question

If \(3.00 \mathrm{~L}\) of sulfur dioxide gas reacts with \(1.25 \mathrm{~L}\) of oxygen gas, what is the volume of sulfur trioxide gas produced? Assume all gases are at the same temperature and pressure. $$\mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{3}(g)$$

Short answer

2.50 L of sulfur trioxide gas is produced.

Step-by-step solution

Write the Balanced Chemical Equation

First, let's write the balanced chemical equation for the reaction. The reaction of sulfur dioxide with oxygen to form sulfur trioxide is given by: \[ 2 ext{SO}_2(g) + ext{O}_2(g) ightarrow 2 ext{SO}_3(g) \] This equation shows that 2 volumes of \( ext{SO}_2 \) react with 1 volume of \( ext{O}_2 \) to form 2 volumes of \( ext{SO}_3 \).

Determine Limiting Reactant

We need to determine which reactant limits the reaction. According to the balanced equation, the stoichiometric ratio of \( ext{SO}_2 \) to \( ext{O}_2 \) is 2:1. Given, we have 3.00 L of \( ext{SO}_2 \) and 1.25 L of \( ext{O}_2 \), the required amount of \( ext{O}_2 \) is \( \frac{3.00 ext{ L}}{2} = 1.50 ext{ L} \). Only 1.25 L of \( ext{O}_2 \) is available, so \( ext{O}_2 \) is the limiting reactant.

Calculate the Volume of Sulfur Trioxide Produced

Since \( ext{O}_2 \) is the limiting reactant, it determines the amount of \( ext{SO}_3 \) produced. The balanced equation shows that 1 volume of \( ext{O}_2 \) produces 2 volumes of \( ext{SO}_3 \). Therefore, 1.25 L of \( ext{O}_2 \) will produce \( 2 \times 1.25 ext{ L} = 2.50 ext{ L} \) of \( ext{SO}_3 \).

Key concepts

Balanced Chemical Equation

In a chemical reaction, a balanced equation represents the quantitative relationship between the reactants and products. For the reaction of sulfur dioxide (\(\mathrm{SO}_2(g)\)) with oxygen (\(\mathrm{O}_2(g)\)) to form sulfur trioxide (\(\mathrm{SO}_3(g)\)), the balanced equation is:\[2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g)\]This balanced chemical equation indicates that two molecules of sulfur dioxide react with one molecule of oxygen to produce two molecules of sulfur trioxide. Balanced equations are crucial because they obey the law of conservation of mass, meaning the number of each type of atom is the same on both sides of the equation.

• The coefficients in a balanced chemical equation reflect the ratios in which substances react.
• For gases, these coefficients can also represent the volume ratio when measured under the same conditions of temperature and pressure.

When dealing with problems involving gas volumes, keeping the concept of a balanced equation in mind helps predict how much product will form or how much of each reactant is needed.

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that gets consumed first and prevents the reaction from continuing. In other words, it limits the amount of product that can be formed. For our exercise, to determine the limiting reactant, you compare the mole ratio from the balanced equation to the actual amount of reactants you have.
Let's see this in action:

• Based on the balanced equation, 2 volumes of \(\mathrm{SO}_2\) are needed for every 1 volume of \(\mathrm{O}_2\).
• We have 3.00 L of \(\mathrm{SO}_2\) and 1.25 L of \(\mathrm{O}_2\).
• To fully react with 3.00 L of \(\mathrm{SO}_2\), we would need \(\frac{3.00 \text{ L of } \mathrm{SO}_2}{2} = 1.50 \text{ L of } \mathrm{O}_2\).
• Since only 1.25 L of \(\mathrm{O}_2\) is available, \(\mathrm{O}_2\) is the limiting reactant.

It is essential to always identify the limiting reactant because it will directly affect the amount of product produced in the reaction.

Volume of Gas

When all reactants and products in a chemical reaction are gases, the Balanced Chemical Equation can be directly related to their volumes. This is possible because, under constant temperature and pressure, gases follow the ideal gas law. In our case, where all gases react under the same conditions, the volume ratios correspond to the mole ratios given by the balanced equation.
For the formation of sulfur trioxide, the balanced equation tells us:

• 1 volume of \(\mathrm{O}_2\) produces 2 volumes of \(\mathrm{SO}_3\).

Given that \(\mathrm{O}_2\) is the limiting reactant with only 1.25 L available, we calculate the volume of sulfur trioxide (\(\mathrm{SO}_3\)) as follows:

• Since 1 volume of \(\mathrm{O}_2\) results in 2 volumes of \(\mathrm{SO}_3\), simply multiply: \(2 \times 1.25 \text{ L} = 2.50 \text{ L of } \mathrm{SO}_3\)

Therefore, 2.50 L of sulfur trioxide gas will be produced when 1.25 L of oxygen gas is fully consumed.