Problem 32 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

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 furnace such that the contents undergo self- reduction. The weight (in \(\mathrm{kg}\) ) of Pb produced per \(\mathrm{kg}\) of \(\mathrm{O}_{2}\) consumed is (Atomic weights in \(\mathrm{g} \mathrm{mol}^{-1}: \mathrm{O}=16, \mathrm{~S}=32, \mathrm{~Pb}=207\) ) [Adv. 2018]

Step-by-Step Solution

Answer

The weight of Pb produced per kg of \(\text{O}_2\) consumed is 4.31 kg.

Step 1: Write the Chemical Equations

The ore Galena is lead sulfide ( formule: PbS) and is oxidized by air. This process involves two reactions:\[\text{2PbS + 3O}_2 \rightarrow \text{2PbO + 2SO}_2\] This is oxidation of PbS to PbO.\[\text{PbO + PbS} \rightarrow \text{2Pb + SO}_2\] This is self-reduction of PbO to Pb.

Step 2: Calculate Oxygen Consumption in Oxidation

In the oxidation reaction:\[\text{2PbS + 3O}_2 \rightarrow \text{2PbO + 2SO}_2\]To oxidize 2 moles of PbS, we require 3 moles of oxygen (\(\text{O}_2\)). Therefore, each mole of oxygen results in \(\frac{2}{3}\) moles of products containing lead oxide (\(\text{PbO}\)).

Step 3: Calculate Lead Production from PbO

In the reduction reaction: \[\text{PbO + PbS} \rightarrow \text{2Pb + SO}_2\]\(\text{Each mole of PbO produces 1 mole of Pb}\). Thus, from \(\frac{2}{3}\) moles of PbO, we get \(\frac{2}{3}\) moles of Pb.

Step 4: Calculate Total Moles of Pb Produced

From Step 2 and Step 3, per mole of \(\text{O}_2\), we produce \(\frac{2}{3}\) moles of Pb.

Step 5: Convert Moles of Pb to Mass of Pb

\(\text{The atomic weight of lead (Pb) is 207 g/mol.}\)\(\text{Conversion of moles of Pb to mass of Pb:}\)\[\frac{2}{3} \text{ moles} \times 207 \text{ g/mol} = 138 \text{ g}\]Thus, the weight of Pb produced per mole of \(\text{O}_2\) consumed is 138 g or 0.138 kg.

Step 6: Weight of \(\text{O}_2\)

The molecular weight of \(\text{O}_2\) is \(32\) g/mol. Thus, 1 kg of \(\text{O}_2\) is equivalent to \[\frac{1000\text{ g}}{32 \text{ g/mol}} = 31.25 \text{ moles of \(\text{O}_2\)}\].

Key Concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, often called redox reactions, are a type of chemical reaction where the oxidation states of participating elements change. These reactions are vital in processes such as energy production and various industrial processes, including ore refinement. In a redox reaction, one reactant undergoes oxidation by losing electrons, while the other reactant undergoes reduction by gaining electrons.

• Oxidation: This is the process where a molecule, atom, or ion loses electrons. For example, in the case of galena (PbS), lead sulfide undergoes oxidation, reacting with oxygen to form lead oxide (PbO).
• Reduction: The opposite of oxidation, reduction involves the gain of electrons. In the reaction involving galena, PbO undergoes reduction by reacting with remaining lead sulfide (PbS) to form elemental lead (Pb).

Understanding redox reactions helps to trace the flow of electrons and understand the changes in energy and material transformations during the reactions. In stoichiometry, these reactions are balanced by equalizing the number of electron transfers between oxidation and reduction.

Atomic Weights

Atomic weights, also known as atomic masses, are essential for stoichiometry. Knowing the atomic weights allows us to convert between moles of a substance and grams of that substance. This conversion is crucial for calculating the yields of a reaction from the reactants used.

• Importance of Atomic Weights: In reactions, the masses involved need to be known to predict the products accurately. For example, with galena, the atomic weights of oxygen (16 g/mol), sulfur (32 g/mol), and lead (207 g/mol) are used to convert moles into grams.
• Practical Calculation: In the given exercise, calculating the mass of Pb produced involves using its atomic weight to determine how many grams result from reaction stoichiometry.

The concept of atomic weights is integral to translating chemical equations into quantitative data, which then supports laboratory or industrial scale production.

Chemical Equations

Chemical equations represent a chemical reaction with symbols and formulas, showing both the reactants and products. They are balanced to reflect the conservation of mass and charge.

• Balancing Equations: In order to comply with the law of conservation of mass, the total atoms of each element must be the same on both sides of the equation. This is seen in the reaction, \(2\text{PbS} + 3\text{O}_2 \rightarrow 2\text{PbO} + 2\text{SO}_2\), where reactants correctly balance with products.
• Types of Chemical Equations: Equations may be simple or complex, often involving multiple steps. Here, the oxidation and self-reduction reactions are chained to derive elemental lead from galena.

Chemical equations not only describe the quantifiable aspects of reactions but also provide insights into the stoichiometric relationships necessary for real-world applications.

Oxygen Consumption

Oxygen consumption is a critical aspect in oxidation reactions and plays a significant role in determining the quantity of products formed in stoichiometric calculations. During oxidation, galena reacts with oxygen, which is plentiful in the air, to transform into lead oxide and sulfur dioxide.

• Role in Reactions: Oxygen acts as an oxidizing agent and it is crucial to know the amount consumed to predict the yield of lead. In the exercise, it was indicated that for every 3 moles of \(\text{O}_2\) used, 2 moles of lead oxide are formed.
• Calculating Consumption: To find out how much lead is produced per kilogram of oxygen consumed, one must understand the stoichiometric proportion of oxygen used as given by the balanced chemical equations.

Understanding oxygen consumption not only aids in calculating the mass of products formed but also helps in planning industrial processes where oxygen is a reactant, anticipating how much material will be used.