Problem 12 - Chemistry: An Atoms-Focused Approach

Question

Combustion of sulfur may, depending on reaction conditions, produce \(\mathrm{SO}_{2}\) or \(\mathrm{SO}_{3}\). If \(x\) grams of \(\mathrm{O}_{2}\) combine with \(y\) grams of sulfur to form \(\mathrm{SO}_{2},\) how many grams of \(\mathrm{O}_{2}\) combine with \(y\) grams of sulfur to form \(\mathrm{SO}_{3} ?\)

Step-by-Step Solution

Answer

Answer: (3/2) * (y * 32.00 / 32.06) grams of oxygen.

Step 1: Write the balanced equations for the formation of SO₂ and SO₃

The balanced equations for the formation of SO₂ and SO₃ are as follows: S + O₂ → SO₂ S + 3/2 O₂ → SO₃

Step 2: Convert grams to moles

To work with amounts of reactants in chemical equations, we need to convert grams to moles using the molar mass of each element. Molar mass of S = 32.06 g/mol Molar mass of O₂ = 32.00 g/mol So, for SO₂: moles of S = y g / 32.06 g/mol moles of O₂ = x g / 32.00 g/mol And for SO₃: We'll use the same moles of S as before, because it's the same amount of sulfur involved in both reactions.

Step 3: Find the moles of O₂ needed for SO₃ formation

To find the moles of O₂ needed to form SO₃, we'll use the mole-to-mole ratio from the balanced equation: 1 S : 3/2 O₂ moles of O₂ for SO₃ = moles of S (y/32.06) * (3/2)

Step 4: Convert moles of O₂ back to grams

Now we'll convert the moles of O₂ needed for SO₃ formation back to grams: grams of O₂ for SO₃ = moles of O₂ for SO₃ * 32.00 g/mol grams of O₂ for SO₃ = (y/32.06) * (3/2) * 32.00

Step 5: Simplify the expression

Finally, we'll simplify the expression for the grams of O₂ needed to form SO₃: grams of O₂ for SO₃ = (3/2) * (y * 32.00 / 32.06) So, when y grams of sulfur combine with x grams of O₂ to form SO₂, (3/2) * (y * 32.00 / 32.06) grams of O₂ combine with y grams of sulfur to form SO₃.

Key Concepts

Understanding Combustion Reactions

In a combustion reaction, a substance combines with oxygen to produce heat and often light, resulting in the formation of new substances such as gases or liquids. These reactions are a fundamental type of chemical reaction that involve burning.
In our example, sulfur combusts with oxygen to produce either sulfur dioxide (\(\text{SO}_2\)) or sulfur trioxide (\(\text{SO}_3\)).
Key features of combustion reactions include:

• They always involve oxygen as one of the reactants.
• They release energy, usually in the form of heat and light.
• The products are generally oxides, like \(\text{SO}_2\) and \(\text{SO}_3\).

In our context, the choice between forming \(\text{SO}_2\) or \(\text{SO}_3\) depends on specific conditions, such as temperature or availability of oxygen. Recognizing how different combustion conditions affect product formation is crucial for understanding these chemical processes.

Exploring Oxidation States

Oxidation states, also known as oxidation numbers, are a way to keep track of electrons in a chemical reaction. They help us understand how electrons are transferred between atoms, which is central to reactions involving combustion.
In sulfur combustion reactions, you'll come across these oxidation states:

• Sulfur (S) starts in its elemental form, with an oxidation state of 0.
• In \(\text{SO}_2\), sulfur's oxidation state increases to +4.
• In \(\text{SO}_3\), the oxidation state of sulfur further increases to +6.

The increase in oxidation state from \(\text{SO}_2\) to \(\text{SO}_3\) reflects the gain of more oxygen atoms, meaning more electrons are being "pulled" away from sulfur, indicating its higher oxidation number. Understanding the concept of oxidation states is vital, especially in combustion reactions, as it allows scientists to predict how different species react.

Importance of Balancing Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They are balanced to adhere to the law of conservation of mass, which states that mass cannot be created or destroyed.
A balanced chemical equation ensures that the number of each type of atom is the same on both sides of the equation. Let's look at these sulfur combustion equations:

• For \(\text{SO}_2\): \(\text{S} + \text{O}_2 \rightarrow \text{SO}_2\), one sulfur and two oxygens react to form sulfur dioxide.
• For \(\text{SO}_3\): \(\text{S} + \frac{3}{2}\text{O}_2 \rightarrow \text{SO}_3\), indicating that one sulfur reacts with one and a half \(\text{O}_2\) molecules to form sulfur trioxide.

Balancing is crucial because it accurately reflects the stoichiometry of the reaction, ensuring that predictions and calculations, such as determining reactant quantities, are correct. This step is a foundational aspect of solving problems in chemistry and allows you to convert between moles and grams accurately, as demonstrated in the exercise provided.