Write balanced equations for the sfollowing reactions:
Lead sulphide is heating in air.

Answer

To write the balanced equation for the reaction of lead sulfide (PbS) when heated in air, we follow these steps:

### Step 1: Identify the Reactants and Products
- The reactant is lead sulfide (PbS).
- The products formed when PbS is heated in air (which contains oxygen) are lead(II) oxide (PbO) and sulfur dioxide (SO2).

### Step 2: Write the Unbalanced Equation
The unbalanced equation can be written as:
\[ \text{PbS} + \text{O}_2 \rightarrow \text{PbO} + \text{SO}_2 \]

### Step 3: Balance the Equation
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides.

1. **Count the atoms:**
- Left side:
- Pb: 1
- S: 1
- O: 2 (from O2)
- Right side:
- Pb: 1
- S: 1
- O: 3 (1 from PbO and 2 from SO2)

2. **Balancing Oxygen:**
- We have 3 oxygen atoms on the right side and 2 on the left. To balance the oxygen, we can adjust the coefficients.
- If we use 3/2 O2, we will have 3 oxygen atoms on the left.

3. **Adjust the coefficients:**
- The balanced equation now looks like:
\[ \text{PbS} + \frac{3}{2} \text{O}_2 \rightarrow \text{PbO} + \text{SO}_2 \]

4. **Eliminate the fraction by multiplying through by 2:**
- This gives:
\[ 2 \text{PbS} + 3 \text{O}_2 \rightarrow 2 \text{PbO} + 2 \text{SO}_2 \]

### Final Balanced Equation
The final balanced equation for the reaction of lead sulfide when heated in air is:
\[ 2 \text{PbS} + 3 \text{O}_2 \rightarrow 2 \text{PbO} + 2 \text{SO}_2 \]