Use equation `2H_(2)O(l) to 2H_(2)(g)` to answer the following
What volume of `O_(2)` will be produced if the volume of `H_(2)` produced is `2500 cm^(3)` under similar conditions?

Explanation

To solve the problem, we start with the balanced chemical equation:

\[ 2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \]

From this equation, we can see the following relationships:
- 2 moles (or volumes) of water produce 2 moles (or volumes) of hydrogen gas and 1 mole (or volume) of oxygen gas.

### Step-by-Step Solution:

1. **Identify the relationship from the balanced equation**:
From the equation, we see that:
- 2 volumes of \( H_2 \) are produced for every 1 volume of \( O_2 \).

2. **Set up the known volume of \( H_2 \)**:
We are given that the volume of \( H_2 \) produced is \( 2500 \, cm^3 \).

3. **Determine the volume of \( O_2 \)**:
Since 2 volumes of \( H_2 \) produce 1 volume of \( O_2 \), we can set up the ratio:
\[ \frac{Volume \, of \, H_2}{Volume \, of \, O_2} = 2:1 \]
Therefore, if the volume of \( H_2 \) is \( 2500 \, cm^3 \), we can express the volume of \( O_2 \) as:
\[ Volume \, of \, O_2 = \frac{Volume \, of \, H_2}{2} \]

4. **Calculate the volume of \( O_2 \)**:
Plugging in the known volume of \( H_2 \):
\[ Volume \, of \, O_2 = \frac{2500 \, cm^3}{2} = 1250 \, cm^3 \]

5. **Final Answer**:
The volume of \( O_2 \) produced is \( 1250 \, cm^3 \).

### Summary:
The volume of \( O_2 \) produced when \( 2500 \, cm^3 \) of \( H_2 \) is generated is \( 1250 \, cm^3 \).

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