If 30 ml of `H_2` and 20 ml of `O_2` react to form water, what is left at the end of the reaction ?

Options

A. 10 ml of `H_2`
B. 5 ml of `H_2`
C. 10 ml of `O_2`
D. 5 ml of `O_2`

Answer

To solve the problem of what is left after the reaction of 30 ml of \( H_2 \) and 20 ml of \( O_2 \) to form water, we can follow these steps:

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

### Step 2: Determine the Stoichiometric Ratios
From the balanced equation, we can see that:
- 2 volumes of \( H_2 \) react with 1 volume of \( O_2 \).
- Therefore, 1 volume of \( H_2 \) reacts with 0.5 volumes of \( O_2 \).

### Step 3: Calculate the Required Amount of \( O_2 \) for the Given Amount of \( H_2 \)
Given that we have 30 ml of \( H_2 \):
- The amount of \( O_2 \) required to react with 30 ml of \( H_2 \) is:
\[ 30 \, \text{ml} \, H_2 \times \frac{0.5 \, \text{ml} \, O_2}{1 \, \text{ml} \, H_2} = 15 \, \text{ml} \, O_2 \]

### Step 4: Compare the Available \( O_2 \) with the Required \( O_2 \)
We have 20 ml of \( O_2 \) available, but only 15 ml is needed to react with all of the 30 ml of \( H_2 \). This means that \( H_2 \) is the limiting reagent because it will be completely consumed.

### Step 5: Calculate the Remaining Amount of \( O_2 \)
After the reaction, the amount of \( O_2 \) left can be calculated as follows:
\[ \text{Initial } O_2 = 20 \, \text{ml} - \text{Consumed } O_2 = 20 \, \text{ml} - 15 \, \text{ml} = 5 \, \text{ml} \]

### Step 6: Calculate the Amount of Water Produced
From the balanced equation, 2 volumes of \( H_2 \) produce 2 volumes of \( H_2O \). Therefore, 30 ml of \( H_2 \) will produce:
\[ 30 \, \text{ml} \, H_2 \rightarrow 30 \, \text{ml} \, H_2O \]

### Final Result
At the end of the reaction, the amounts left are:
- \( H_2 \): 0 ml (completely consumed)
- \( O_2 \): 5 ml (remaining)
- \( H_2O \): 30 ml (produced)

### Conclusion
The final amounts after the reaction are:
- 5 ml of \( O_2 \) is left at the end of the reaction.