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.
What gas if any will be present on cooling to room temperature?

Explanation

To solve the problem, we need to determine how much of each reactant (hydrogen and oxygen) is consumed in the reaction to form water and what remains after the reaction when the mixture is cooled to room temperature.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The balanced equation for the reaction of hydrogen and oxygen to form water is:
\[ 2H_2 + O_2 \rightarrow 2H_2O \]

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

3. **Determine the Stoichiometric Ratios**:
According to the balanced equation:
- 2 volumes of \( H_2 \) react with 1 volume of \( O_2 \).
This means:
- 22.4 L of \( H_2 \) would require \( \frac{22.4}{2} = 11.2 \, L \) of \( O_2 \).
- Conversely, 22.4 L of \( O_2 \) would require \( 2 \times 22.4 = 44.8 \, L \) of \( H_2 \).

4. **Identify the Limiting Reagent**:
- Since we only have 22.4 L of \( H_2 \), it can react with only 11.2 L of \( O_2 \).
- The available \( O_2 \) is 22.4 L, which means there is excess \( O_2 \).
- Therefore, \( H_2 \) is the limiting reagent because it will be completely consumed in the reaction.

5. **Calculate the Remaining Volume of \( O_2 \)**:
- Since \( 22.4 \, L \) of \( H_2 \) will consume \( 11.2 \, L \) of \( O_2 \), the remaining volume of \( O_2 \) after the reaction will be:
\[ \text{Remaining } O_2 = 22.4 \, L - 11.2 \, L = 11.2 \, L \]

6. **Determine the State of Water**:
- At 100°C, water is in the gaseous state (steam).
- Upon cooling to room temperature, water will condense into liquid form.

7. **Conclusion**:
- After the reaction and cooling, the only gas present will be the unreacted \( O_2 \).
- Therefore, the answer is:
\[ \text{11.2 L of } O_2 \text{ will be present as gas on cooling.} \]

### Final Answer:
**11.2 L of \( O_2 \) will be present on cooling to room temperature.**

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