Hydrogen and oxygen combine to form water according to the following equation:
`2H_(2)+O_(2) to 2H_(2)O`. A mixture of 22.4 L of `H_(2)` and 22.4 L of `O_(2)` at `100^(@)C` is ignited.
Calculate the volume of steam produced.

Explanation

To solve the problem of calculating the volume of steam produced when hydrogen and oxygen combine to form water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**: The balanced chemical equation for the reaction is:
\[ 2H_2 + O_2 \rightarrow 2H_2O \]

2. **Determine the Initial Volumes**: We are given:
- Volume of \( H_2 = 22.4 \, L \)
- Volume of \( O_2 = 22.4 \, L \)

3. **Convert Volumes to Moles**: At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 L. Therefore, we can calculate the moles of each gas:
\[ \text{Moles of } H_2 = \frac{22.4 \, L}{22.4 \, L/mol} = 1 \, mol \]
\[ \text{Moles of } O_2 = \frac{22.4 \, L}{22.4 \, L/mol} = 1 \, mol \]

4. **Identify the Limiting Reagent**: According to the stoichiometry of the reaction:
- 2 moles of \( H_2 \) react with 1 mole of \( O_2 \).
- From the moles of \( H_2 \) (1 mol), we can find the amount of \( O_2 \) needed:
\[ \text{Moles of } O_2 \text{ required} = \frac{1 \, mol \, H_2}{2} = 0.5 \, mol \]
- Since we have 1 mol of \( O_2 \) available, which is greater than 0.5 mol, \( H_2 \) is the limiting reagent.

5. **Calculate Moles of Water Produced**: According to the balanced equation:
- 2 moles of \( H_2 \) produce 2 moles of \( H_2O \).
- Therefore, 1 mole of \( H_2 \) will produce 1 mole of \( H_2O \).

6. **Convert Moles of Water to Volume**: The volume of steam produced can be calculated using the molar volume at STP:
\[ \text{Volume of } H_2O = \text{Moles of } H_2O \times \text{Molar Volume} \]
\[ \text{Volume of } H_2O = 1 \, mol \times 22.4 \, L/mol = 22.4 \, L \]

### Final Answer:
The volume of steam produced is **22.4 liters**.

---