JEE Challenger
More from General Principles and Processes of Isolation of Elements
ChemMedAdv 2018 • Paper 2

Mass of Lead Produced Per Unit Oxygen in Self Reduction

Galena (a lead ore) is partially oxidized by passing air through it at elevated temperature. After a certain duration, the flow of air is cut off, but heating is sustained in a closed furnace so that the contents undergo self-reduction. Calculate the mass (in kg\text{kg}) of Pb\text{Pb} generated per kg\text{kg} of O2\text{O}_2 consumed. (Given atomic weights in g mol−1\text{g mol}^{-1}: O=16\text{O} = 16, S=32\text{S} = 32, Pb=207\text{Pb} = 207)

Official Numerical Answer6.47

Topics & Concepts

General Principles and Processes of Isolation of ElementsMetallurgy

Step-by-Step Solution

To determine the mass of lead (Pb\text{Pb}) produced per unit mass of oxygen (O2\text{O}_2) consumed during the self-reduction of Galena (PbS\text{PbS}), we consider the chemical reactions involved.

Step 1: Chemical Reactions for Self-Reduction

During partial oxidation, PbS\text{PbS} is oxidized to PbO\text{PbO} and/or PbSO4\text{PbSO}_4. In the subsequent self-reduction step (without air supply), these oxidized species react with the remaining PbS\text{PbS} to yield metallic lead (Pb\text{Pb}).

Pathway 1: Via Lead(II) Oxide (PbO\text{PbO})

  1. Oxidation step: 2PbS+3O2→2PbO+2SO22\text{PbS} + 3\text{O}_2 \rightarrow 2\text{PbO} + 2\text{SO}_2

  2. Self-reduction step: 2PbO+PbS→3Pb+SO22\text{PbO} + \text{PbS} \rightarrow 3\text{Pb} + \text{SO}_2

Adding the two balanced equations to find the overall reaction: (2PbS+3O2)+(2PbO+PbS)→(2PbO+2SO2)+(3Pb+SO2)(2\text{PbS} + 3\text{O}_2) + (2\text{PbO} + \text{PbS}) \rightarrow (2\text{PbO} + 2\text{SO}_2) + (3\text{Pb} + \text{SO}_2)

Simplifying the net reaction: 3PbS+3O2→3Pb+3SO23\text{PbS} + 3\text{O}_2 \rightarrow 3\text{Pb} + 3\text{SO}_2 ⇒PbS+O2→Pb+SO2\Rightarrow \text{PbS} + \text{O}_2 \rightarrow \text{Pb} + \text{SO}_2

Pathway 2: Via Lead(II) Sulfate (PbSO4\text{PbSO}_4)

  1. Oxidation step: PbS+2O2→PbSO4\text{PbS} + 2\text{O}_2 \rightarrow \text{PbSO}_4

  2. Self-reduction step: PbSO4+PbS→2Pb+2SO2\text{PbSO}_4 + \text{PbS} \rightarrow 2\text{Pb} + 2\text{SO}_2

Adding the two balanced equations: 2PbS+2O2→2Pb+2SO22\text{PbS} + 2\text{O}_2 \rightarrow 2\text{Pb} + 2\text{SO}_2 ⇒PbS+O2→Pb+SO2\Rightarrow \text{PbS} + \text{O}_2 \rightarrow \text{Pb} + \text{SO}_2

Step 2: Mole and Mass Ratio Calculation

In either pathway, the overall stoichiometric relation is: 1 mole of O2 consumed  ⟹  1 mole of Pb produced1 \text{ mole of } \text{O}_2 \text{ consumed} \implies 1 \text{ mole of } \text{Pb} \text{ produced}

Using the given atomic masses:

  • Molar mass of O2=2×16=32 g mol−1\text{O}_2 = 2 \times 16 = 32\text{ g mol}^{-1}
  • Atomic mass of Pb=207 g mol−1\text{Pb} = 207\text{ g mol}^{-1}

The ratio of the mass of Pb\text{Pb} produced to the mass of O2\text{O}_2 consumed is: Mass ratio=Mass of PbMass of O2=207 g32 g=20732\text{Mass ratio} = \frac{\text{Mass of Pb}}{\text{Mass of O}_2} = \frac{207\text{ g}}{32\text{ g}} = \frac{207}{32}

Mass ratio=6.46875 kg of Pb per kg of O2\text{Mass ratio} = 6.46875\text{ kg of Pb per kg of O}_2

Rounding off to two decimal places: 6.476.47

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