To balance the chemical reaction \( xP + 5O_2 \rightarrow yP_2O_5 \), we first need to identify the number of phosphorus (P) and oxygen (O) atoms on both sides of the equation.

1. **Count P and O on the right side (product side)**:
- In \( yP_2O_5 \), there are \( 2y \) phosphorus atoms and \( 5y \) oxygen atoms.

2. **Count P and O on the left side (reactant side)**:
- On the left, we have \( xP \) contributing \( x \) phosphorus atoms.
- The \( 5O_2 \) contributes \( 10 \) oxygen atoms (since \( 5O_2 = 5 \times 2 = 10O \)).

Now we can set up the equations for balancing:

### Balancing Phosphorus:
\[
x = 2y
\]

### Balancing Oxygen:
\[
10 = 5y
\]

From the oxygen balance, we can solve for \( y \):
\[
y = \frac{10}{5} = 2
\]

Now substitute \( y \) back into the phosphorus equation:
\[
x = 2(2) = 4
\]

### Summary:
- \( x = 4 \)
- \( y = 2 \)

### Balanced Equation:
Putting it all together, the balanced chemical equation is:
\[
4P + 5O_2 \rightarrow 2P_2O_5
\]