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?

Short answer

When \(50 L\) of \(SO_2\) and \(25 L\) of \(O_2\) react completely at the same temperature and pressure, \(50 L\) of \(SO_3\) will be produced, according to the stoichiometry of the balanced chemical equation.

Step-by-step solution

Determine the stoichiometric ratio

In the given balanced chemical equation, the stoichiometric ratio between the reactants and the product is \(2 \mathrm{SO}_{2} : 1 \mathrm{O}_{2} : 2 \mathrm{SO}_{3}\). This means for every two moles of \(SO_2\), one mole of \(O_2\) is required to produce two moles of \(SO_3\). We can apply this ratio to volumes as well, since they are directly proportional to moles at the same temperature and pressure.

Identify the limiting reactant

Now, we find the limiting reactant: the reactant that gets consumed first and hence determines the amount of product formed. We have \(50 L\) of \(SO_2\) and \(25 L\) of \(O_2\). According to the stoichiometric ratio, we need two volumes of \(SO_2\) for one volume of \(O_2\). So, for \(25 L\) of \(O_2\), we would need \(2 \times 25 L = 50 L\) of \(SO_2\). Since we have exactly \(50 L\) of \(SO_2\), it is the limiting reactant and both reactants will be completely consumed.

Calculate the volume of SO3 formed

Once we know that the limiting reactant is \(SO_2\), we can determine the volume of \(SO_3\) produced. According to the stoichiometric ratio, for every 2 volumes of \(SO_2\), 2 volumes of \(SO_3\) are produced. Since there are \(50 L\) of \(SO_2\), we can calculate the volume of \(SO_3\) formed as follows: Volume of \(SO_3\) = (Volume of \(SO_2\))/(Volume ratio of \(SO_2\)) × (Volume ratio of \(SO_3\)) = \(\frac{50 L}{2} \times 2 = 50 L\) Hence, \(50 L\) of \(SO_3\) will be produced at the same temperature and pressure when all the \(SO_2\) and \(O_2\) are consumed.

Key concepts

Limiting Reactant

In the study of chemical reactions, understanding the limiting reactant is crucial. The limiting reactant is the substance that is entirely used up in a reaction, which halts the production of any additional products once it is exhausted. Identifying the limiting reactant allows us to predict the maximum amount of product that can be formed from the given reactants.
For this, we compare the amount of each reactant used against the stoichiometric ratio from the balanced equation. In our example with sulfur dioxide ( SO_2 ) and oxygen ( O_2 ), you notice that the stoichiometric ratio is 2 volumes of SO_2 for every 1 volume of O_2 . Given 50 L of SO_2 and 25 L of O_2 , they match the ratio perfectly, indicating that both reactants will be completely consumed at the same time. Here, however, SO_2 is identified as the limiting reactant because it precisely controls the amount of SO_3 formed.

Chemical Reactions

Chemical reactions represent the process by which reactants are transformed into products. For the reaction between sulfur dioxide and oxygen, the equation is:\[ 2 SO_2 (g) + O_2 (g) \rightarrow 2 SO_3 (g) \] This balanced chemical equation not only illustrates the substances involved but also reflects their stoichiometric relationships.

• Reactants (SO_2 and O_2): Existing at the start of the reaction.
• Products (SO_3): Newly formed substances.
• Stoichiometric coefficients: Numbers indicating proportions (2:1:2).

These equations follow the law of conservation of mass, meaning the total mass of reactants equals the total mass of products. Knowing this, students can establish the practical aspects of reactions like yields and reactant efficiency.

Gas Laws

Gas laws are fundamental principles that describe the behavior of gases under varying conditions. In our exercise, we can apply the concept that under the same temperature and pressure, equal volumes of gases contain an equal number of molecules, according to Avogadro's Law. This is why the same volumes are used to explore stoichiometric relationships for gaseous reactions.
Under constant conditions, the volumes of reactants and products are proportional to the coefficients in the balanced chemical equation. Hence, 50 L of SO_2 and 25 L of O_2 reacting precisely demonstrates this principle, resulting in a stoichiometric production of 50 L of SO_3 .
Understanding gas laws helps in predicting how gases will react and behave, making them indispensable for solving problems related to gaseous reactions, such as ours. By using these principles, you can easily convert from moles to volumes, allowing a seamless analysis of gaseous reactants and products.