Problem 15 - Chemistry and Chemical Reactivity

Question

A major source of air pollution years ago was the metals industry. One common process involved "roasting" metal sulfides in the air: $$ 2 \mathrm{PbS}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{PbO}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{g}) $$ If you heat 2.5 mol of \(\mathrm{PbS}\) in the air, what amount of \(\mathrm{O}_{2}\) is required for complete reaction? What amounts of \(\mathrm{PbO}\) and \(\mathrm{SO}_{2}\) are expected?

Step-by-Step Solution

Answer

3.75 moles of \( \mathrm{O}_2 \) are needed. 2.5 moles of \( \mathrm{PbO} \) and 2.5 moles of \( \mathrm{SO}_2 \) are produced.

Step 1: Understand the Reaction

The reaction given is \( 2 \ \mathrm{PbS} + 3 \ \mathrm{O}_2 \rightarrow 2 \ \mathrm{PbO} + 2 \ \mathrm{SO}_2 \). We need to find out how much \( \mathrm{O}_2 \) is required to react with 2.5 moles of \( \mathrm{PbS} \). Additionally, we need to determine the amount of \( \mathrm{PbO} \) and \( \mathrm{SO}_2 \) produced.

Step 2: Utilize Stoichiometry

Using the balanced equation, we see that 2 moles of \( \mathrm{PbS} \) react with 3 moles of \( \mathrm{O}_2 \). Therefore, the stoichiometric ratio between \( \mathrm{PbS} \) and \( \mathrm{O}_2 \) is 2:3.

Step 3: Calculate \( \mathrm{O}_2 \) Required

Since 2 moles of \( \mathrm{PbS} \) require 3 moles of \( \mathrm{O}_2 \), 2.5 moles of \( \mathrm{PbS} \) will require \( \left( \frac{3}{2} \times 2.5 \right) = 3.75 \) moles of \( \mathrm{O}_2 \).

Step 4: Calculate \( \mathrm{PbO} \) Produced

According to the balanced equation, 2 moles of \( \mathrm{PbS} \) produce 2 moles of \( \mathrm{PbO} \) in a 1:1 ratio. Therefore, 2.5 moles of \( \mathrm{PbS} \) will produce 2.5 moles of \( \mathrm{PbO} \).

Step 5: Calculate \( \mathrm{SO}_2 \) Produced

Similarly, 2 moles of \( \mathrm{PbS} \) produce 2 moles of \( \mathrm{SO}_2 \). Hence, 2.5 moles of \( \mathrm{PbS} \) will produce 2.5 moles of \( \mathrm{SO}_2 \) as well.

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is a foundation for understanding reactions in chemistry. It represents the chemical transformation that occurs, showing both reactants and products. One of the key things about a balanced equation is that it respects the Law of Conservation of Mass. This means the number of atoms for each element is the same on both sides of the equation.
For example, in the reaction given:

• 2 moles of lead sulfide (\( \text{PbS} \)) react with 3 moles of oxygen (\( \text{O}_2 \)).
• This produces 2 moles of lead(II) oxide (\( \text{PbO} \)) and 2 moles of sulfur dioxide (\( \text{SO}_2 \)).

To verify its balance, count how many of each type of atom there is on both sides. You'll see that there are equal numbers of lead, sulfur, and oxygen atoms on both sides, ensuring the reaction is balanced.
This balance is crucial as it allows us to use stoichiometry to predict the outcomes of reactions, like determining the amount of products formed or reactants needed.

Mole-to-Mole Conversions

Mole-to-mole conversions are an essential part of stoichiometry. They allow us to predict the quantities of substances that participate in a chemical reaction, based on the coefficients of a balanced equation. It's like a recipe, where ingredients (reactants) and products are measured in moles, which is a chemical amount.
For the given reaction:

• The equation tells us that 2 moles of \( \mathrm{PbS} \) react with 3 moles of \( \mathrm{O}_2 \).
• From this, we learn that for every 2 moles of \( \mathrm{PbS} \), we need 3 moles of \( \mathrm{O}_2 \) for the reaction to go to completion.

If we have 2.5 moles of \( \mathrm{PbS} \), we can use the stoichiometric relationship \(\left( \frac{3}{2} \right)\), meaning we will need 3.75 moles of \( \mathrm{O}_2 \) to fully react with the \( \mathrm{PbS} \). This type of conversion is necessary to decide how much of each reactant is required and how much of each product is going to be formed after the reaction.

Reaction Yield

Reaction yield is a measure of how much product is actually gained from a reaction compared to what is theoretically possible. In an ideal world, all reactions would proceed with 100% efficiency, meaning the actual yield matches the theoretical yield based on stoichiometry. However, in reality, side reactions, measurement errors, and incomplete reactions can reduce the yield.
For the exercise provided:

• The ideal or theoretical yields for \( \mathrm{PbO} \) and \( \mathrm{SO}_2 \) are calculated using the balanced equation.
• According to the stoichiometry, from 2.5 moles of \( \mathrm{PbS} \), we expect to get 2.5 moles of \( \mathrm{PbO} \) and 2.5 moles of \( \mathrm{SO}_2 \).

We assume all reactants convert completely into products, which rarely happens in laboratory or industrial processes. A comparison of theoretical yield vs. actual yield offers insight into the efficiency of a reaction, helps identify potential issues, and indicates where improvements can be made.