In order to balance the following chemical equation, the values of the coefficients x and y respectively are :

$x\text{Pb(NO}_3)_2 \xrightarrow{\text{Heat}} 2\text{PbO} + y\text{NO}_2 + \text{O}_2$

To balance the given chemical equation:

$$x\text{Pb(NO}_3)_2 \xrightarrow{\text{Heat}} 2\text{PbO} + y\text{NO}_2 + \text{O}_2$$

Let us count the atoms of each element on both sides:

1. Lead (Pb):

• On the RHS, we have $2$ atoms of $\text{Pb}$ (in $2\text{PbO}$).
• To balance $\text{Pb}$ on the LHS, we must have $x = 2$.

1. Nitrogen (N):

• With $x = 2$, the LHS has $2 \times 2 = 4$ atoms of $\text{N}$ (from $2\text{Pb(NO}_3)_2$).
• To balance $\text{N}$ on the RHS, we must have $y = 4$ atoms of $\text{N}$ (in $y\text{NO}_2$). Thus, $y = 4$.

1. Oxygen (O):

• Let's verify if oxygen is balanced with $x = 2$ and $y = 4$:
• LHS: $2 \times (3 \times 2) = 12$ oxygen atoms.
• RHS: $2$ (from $2\text{PbO}$) + $4 \times 2$ (from $4\text{NO}_2$) + $2$ (from $\text{O}_2$) = $2 + 8 + 2 = 12$ oxygen atoms.

Both sides are perfectly balanced. Therefore, $x = 2$ and $y = 4$.