1. Understanding the Process

The problem describes the extraction of lead (Pb) from its ore, Galena (PbS), via a self-reduction process. This process involves two main stages:

1. Partial Roasting: Galena is heated in a controlled supply of air. Some of the lead sulfide is converted into lead(II) oxide (PbO) and lead(II) sulfate (PbSO₄).
2. Self-Reduction: The air supply is cut off, and the mixture is heated to a higher temperature. The remaining lead sulfide (PbS) acts as a reducing agent, reducing the PbO and PbSO₄ to molten lead (Pb).

2. Writing the Balanced Chemical Equations

Let's write down the balanced chemical reactions for each step:

Roasting Reactions: $2\text{PbS}(s) + 3\text{O}_2(g) \rightarrow 2\text{PbO}(s) + 2\text{SO}_2(g) \quad \cdots(1)$ $\text{PbS}(s) + 2\text{O}_2(g) \rightarrow \text{PbSO}_4(s) \quad \cdots(2)$

Self-Reduction Reactions: $\text{PbS}(s) + 2\text{PbO}(s) \rightarrow 3\text{Pb}(l) + \text{SO}_2(g) \quad \cdots(3)$ $\text{PbS}(s) + \text{PbSO}_4(s) \rightarrow 2\text{Pb}(l) + 2\text{SO}_2(g) \quad \cdots(4)$

3. Determining the Overall Stoichiometry

To find the relationship between the oxygen consumed and the lead produced, we need to find the overall net reaction. The process can proceed via two pathways, one involving PbO and the other involving PbSO₄.

Pathway 1 (via PbO): We combine reactions (1) and (3) to eliminate the intermediate PbO. Reaction (1) produces 2 moles of PbO, which are consumed in reaction (3).

• Reaction (1): $2\text{PbS} + 3\text{O}_2 \rightarrow 2\text{PbO} + 2\text{SO}_2$
• Reaction (3): $\text{PbS} + 2\text{PbO} \rightarrow 3\text{Pb} + \text{SO}_2$

Adding these two equations gives: $(2\text{PbS} + 3\text{O}_2) + (\text{PbS} + 2\text{PbO}) \rightarrow (2\text{PbO} + 2\text{SO}_2) + (3\text{Pb} + \text{SO}_2)$ $3\text{PbS} + 3\text{O}_2 + 2\text{PbO} \rightarrow 3\text{Pb} + 3\text{SO}_2 + 2\text{PbO}$ Canceling the intermediate $2\text{PbO}$ from both sides and simplifying by dividing by 3: $\text{PbS} + \text{O}_2 \rightarrow \text{Pb} + \text{SO}_2 \quad \cdots(\text{Overall Reaction A})$

Pathway 2 (via PbSO₄): We combine reactions (2) and (4) to eliminate the intermediate PbSO₄. Reaction (2) produces 1 mole of PbSO₄, which is consumed in reaction (4).

• Reaction (2): $\text{PbS} + 2\text{O}_2 \rightarrow \text{PbSO}_4$
• Reaction (4): $\text{PbS} + \text{PbSO}_4 \rightarrow 2\text{Pb} + 2\text{SO}_2$

Adding these two equations gives: $(\text{PbS} + 2\text{O}_2) + (\text{PbS} + \text{PbSO}_4) \rightarrow (\text{PbSO}_4) + (2\text{Pb} + 2\text{SO}_2)$ $2\text{PbS} + 2\text{O}_2 + \text{PbSO}_4 \rightarrow 2\text{Pb} + 2\text{SO}_2 + \text{PbSO}_4$ Canceling the intermediate $\text{PbSO}_4$ from both sides and simplifying by dividing by 2: $\text{PbS} + \text{O}_2 \rightarrow \text{Pb} + \text{SO}_2 \quad \cdots(\text{Overall Reaction B})$

Both pathways lead to the same overall stoichiometric relationship. Therefore, for the complete process, 1 mole of O₂ is consumed for every 1 mole of Pb produced.

4. Stoichiometric Calculation

From the overall reaction, we have the molar ratio: $1 \text{ mole of } \text{O}_2 \text{ produces } 1 \text{ mole of } \text{Pb}$

Now, we use the given atomic weights to find the molar masses:

• Atomic weight of O = 16 g/mol, so Molar mass of O₂ = $2 \times 16 = 32$ g/mol.
• Atomic weight of Pb = 207 g/mol, so Molar mass of Pb = 207 g/mol.

This means: $32 \text{ g of } \text{O}_2 \text{ produces } 207 \text{ g of } \text{Pb}$

The question asks for the weight (in kg) of Pb produced per kg of O₂ consumed. This is a mass ratio, which is independent of the units used (as long as they are consistent). $\frac{\text{Weight of Pb produced}}{\text{Weight of O}_2 \text{ consumed}} = \frac{207 \text{ g}}{32 \text{ g}} = \frac{207}{32}$

$\frac{207}{32} = 6.46875$

So, for every 1 kg of O₂ consumed, 6.46875 kg of Pb is produced.

5. Final Answer

The weight of Pb produced per kg of O₂ consumed is 6.46875 kg. Rounding this to two decimal places gives 6.47.

Final Answer = 6.47