To solve the problem of determining the maximum theoretical decrease in volume when a mixture of oxygen and hydrogen reacts to form water, we can follow these steps:
### Step-by-Step Solution:
1. **Identify the Initial Volumes:**
- Volume of oxygen (O₂) = 100 mL
- Volume of hydrogen (H₂) = 500 mL
- Total initial volume = Volume of O₂ + Volume of H₂ = 100 mL + 500 mL = 600 mL
2. **Write the Balanced Chemical Equation:**
- The balanced equation for the reaction is:
\[
2H_2 + O_2 \rightarrow 2H_2O
\]
- This indicates that 2 volumes of hydrogen react with 1 volume of oxygen to produce 2 volumes of water.
3. **Determine the Limiting Reactant:**
- According to the stoichiometry of the reaction, 2 volumes of H₂ are required for every 1 volume of O₂.
- For 100 mL of O₂, the required volume of H₂ is:
\[
100 \, \text{mL O₂} \times \frac{2 \, \text{mL H₂}}{1 \, \text{mL O₂}} = 200 \, \text{mL H₂}
\]
- Since we have 500 mL of H₂ available, H₂ is in excess and O₂ is the limiting reactant.
4. **Calculate the Volume of Water Formed:**
- From the balanced equation, 100 mL of O₂ will react with 200 mL of H₂ to produce:
\[
200 \, \text{mL of H₂O} \text{ (as water is in vapor form)}
\]
5. **Calculate the Volume of Gases After Reaction:**
- After the reaction, the volume of gases left will be:
\[
\text{Initial volume} - \text{Volume of gases reacted}
\]
- The volume of gases reacted is the sum of the volumes of O₂ and H₂ that reacted:
\[
\text{Volume reacted} = 100 \, \text{mL O₂} + 200 \, \text{mL H₂} = 300 \, \text{mL}
\]
- Therefore, the remaining volume of gases is:
\[
600 \, \text{mL (initial)} - 300 \, \text{mL (reacted)} = 300 \, \text{mL}
\]
6. **Determine the Maximum Theoretical Decrease in Volume:**
- The maximum theoretical decrease in volume is equal to the volume of gases that reacted:
\[
\text{Decrease in volume} = 300 \, \text{mL}
\]
### Final Answer:
The maximum theoretical decrease in volume at 25°C is **300 mL**.
---
