When heating white lead then find out released gas `(A)` and `(B)`
`underset("white lead")(2PbCO_(3)).Pb(OH)_(2)--overset(Delta)--> Pb_(3)O_(4) + (A) +(B) +H_(2)O`

Options

A. `CO, O_(2)`
B. `CO_(2), O_(2)`
C. `CO_(2), CH_(4)`
D. `CO, CO_(2)`

Explanation

To solve the problem of identifying the gases (A) and (B) released when heating white lead, we can follow these steps:

### Step 1: Write the chemical equation for the decomposition of white lead.
White lead is represented as \(2 \text{PbCO}_3 \cdot \text{Pb(OH)}_2\). When it is heated, it decomposes to form lead(IV) oxide (\(\text{Pb}_3\text{O}_4\)), water (\(H_2O\)), and gases.

### Step 2: Identify the products of the decomposition.
From the video transcript, we know that when white lead is heated, it produces:
- Lead(IV) oxide: \(\text{Pb}_3\text{O}_4\)
- Water: \(H_2O\)
- Two gases which we need to identify as (A) and (B).

### Step 3: Determine the gases released.
The decomposition of lead carbonate (\(\text{PbCO}_3\)) typically results in the release of carbon dioxide (\(\text{CO}_2\)) and carbon monoxide (\(\text{CO}\)).

### Step 4: Write the balanced equation.
The balanced equation for the decomposition can be represented as:
\[ 2 \text{PbCO}_3 \cdot \text{Pb(OH)}_2 \xrightarrow{\Delta} \text{Pb}_3\text{O}_4 + \text{CO}_2 + \text{CO} + H_2O \]
From this equation, we can see that the gases released are:
- Gas (A): Carbon dioxide (\(\text{CO}_2\))
- Gas (B): Carbon monoxide (\(\text{CO}\))

### Conclusion:
The gases released when heating white lead are:
- (A) Carbon dioxide (\(\text{CO}_2\))
- (B) Carbon monoxide (\(\text{CO}\))