Animated Solution for Chemistry - Metallurgy: Galena (an ore) is partially oxidized by passing air through it at high temperature. After some time, the passage of air is stopped, but the heating is continued in a closed furnance such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O${}_2$ consumed is ______ . (Atomic weights in g mol${}^{-1}$ : O = 16, S = 32, Pb = 207)

Imagine you are an industrial chemist tasked with extracting pure lead from its natural ore, Galena ($\text{PbS}$). You could use expensive reducing agents, but nature offers a far more elegant and cost-effective solution: Self-Reduction. This process is a beautiful dance of chemistry where the ore itself acts as the reducing agent. Let's break down the exact mechanics of this reaction and see how we can mathematically track the yield of pure lead.

The Roasting Process

The journey begins in a reverberatory furnace. We take the Galena ore and heat it strongly in the presence of a controlled supply of air. This process is known as roasting. The oxygen in the air reacts with the lead sulfide to form lead oxide and sulfur dioxide gas.

The balanced chemical equation for this step is:

Notice the stoichiometry here. For every two moles of lead sulfide, we require three moles of oxygen gas. However, we don't want to oxidize all the lead sulfide. We only want to oxidize a portion of it. Why? Because the unreacted lead sulfide is the secret weapon for the next step.

The Magic of Self-Reduction

Once a sufficient amount of lead oxide ($\text{PbO}$) has been formed, we completely cut off the air supply. The furnace is now closed, but the intense heating continues.

Now, a fascinating internal reaction occurs. The newly formed lead oxide reacts with the remaining, unoxidized lead sulfide. In this step, the sulfide ion acts as a reducing agent, stripping the oxygen away from the lead oxide to form more sulfur dioxide gas, leaving behind pure, molten lead.

The balanced chemical equation for this self-reduction step is:

To solve our numerical problem, we need a direct relationship between the oxygen we pumped in and the pure lead we got out. We can find this by combining our two equations.

Let's add the roasting equation and the self-reduction equation together:

Notice that the $2\text{PbO}$ appears on both sides of the equation. It acts as an intermediate and cancels out perfectly. Simplifying the equation, we get:

Dividing the entire equation by 3 yields our beautifully simple Master Equation:

This tells us something incredibly powerful: The molar ratio of oxygen consumed to lead produced is exactly 1:1.

The problem asks for the weight of lead produced per $1\text{ kg}$ of oxygen consumed. Let's translate this physical mass into the language of chemistry: moles.

First, we convert the mass of oxygen to grams:

Next, we calculate the moles of oxygen consumed. Remember that oxygen gas is diatomic (${\text{O}}_2$), so its molar mass is $32\text{ g/mol}$:

Thanks to our Master Equation, we know that the moles of lead produced are exactly equal to the moles of oxygen consumed:

Now, we simply convert these moles of lead back into mass. The atomic weight of lead is given as $207\text{ g/mol}$:

Since the question asks for the answer in kilograms, we can conveniently convert the $1000\text{ g}$ factor back into $1\text{ kg}$:

Rounding off to two decimal places, we arrive at our final answer: $6.47\text{ kg}$.

This self-reduction technique is a high-yield concept for competitive exams. Beyond lead, it is also the primary method for extracting copper from copper glance (${\text{Cu}}_2\text{S}$) and mercury from cinnabar ($\text{HgS}$). Always remember to derive the overall stoichiometry to ensure your molar ratios are bulletproof!