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)

Answer

To solve the problem of determining the weight of lead (Pb) produced per kilogram of oxygen (O₂) consumed during the self-reduction of galena (PbS), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Reaction:**
The reaction for the partial oxidation of galena can be represented as:
\[ \text{PbS} + \text{O}_2 \rightarrow \text{Pb} + \text{SO}_2 \]

2. **Convert Kilograms of O₂ to Grams:**
Since we need to work with grams for calculations, convert 1 kg of O₂ to grams:
\[ 1 \text{ kg of O}_2 = 1000 \text{ g of O}_2 \]

3. **Calculate Moles of O₂:**
The molar mass of O₂ is 32 g/mol. To find the number of moles of O₂ in 1000 g:
\[ \text{Moles of O}_2 = \frac{1000 \text{ g}}{32 \text{ g/mol}} = 31.25 \text{ moles} \]

4. **Stoichiometry of the Reaction:**
From the balanced reaction, we see that 1 mole of O₂ produces 1 mole of Pb. Therefore, the moles of Pb produced will also be:
\[ \text{Moles of Pb} = 31.25 \text{ moles} \]

5. **Calculate Mass of Pb Produced:**
The molar mass of Pb is 207 g/mol. To find the mass of Pb produced:
\[ \text{Mass of Pb} = \text{Moles of Pb} \times \text{Molar Mass of Pb} = 31.25 \text{ moles} \times 207 \text{ g/mol} = 6475 \text{ g} \]

6. **Convert Mass of Pb to Kilograms:**
To convert grams to kilograms:
\[ \text{Mass of Pb in kg} = \frac{6475 \text{ g}}{1000} = 6.475 \text{ kg} \]

7. **Final Result:**
Therefore, the weight of Pb produced per kg of O₂ consumed is:
\[ \text{Weight of Pb per kg of O}_2 = 6.475 \text{ kg} \]

### Final Answer:
The weight of Pb produced per kg of O₂ consumed is **6.475 kg**.