If 12.0 lit of `H_(2)` and `8.0` lit of `O_(2)` are allowed ot react, the `O_(2)` left unreacted will be

Options

A. 4.0 lit
B. 6.0 lit
C. 1.0 lit
D. 2.0 lit

Explanation

To solve the problem of how much \( O_2 \) is left unreacted after reacting with \( H_2 \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between hydrogen and oxygen to form water can be represented as:
\[ 2H_2 + O_2 \rightarrow 2H_2O \]
This equation tells us that 2 volumes of hydrogen react with 1 volume of oxygen.

### Step 2: Determine the volume ratio of \( H_2 \) to \( O_2 \)
From the balanced equation, we can see that:
- 2 volumes of \( H_2 \) react with 1 volume of \( O_2 \).
- Therefore, 1 volume of \( H_2 \) will react with \( \frac{1}{2} \) volume of \( O_2 \).

### Step 3: Calculate the amount of \( O_2 \) required for the given \( H_2 \)
Given that we have 12.0 L of \( H_2 \):
\[ \text{Volume of } O_2 \text{ required} = \frac{1}{2} \times \text{Volume of } H_2 = \frac{1}{2} \times 12.0 \text{ L} = 6.0 \text{ L} \]

### Step 4: Determine the initial volume of \( O_2 \)
We are given that we have 8.0 L of \( O_2 \) available for the reaction.

### Step 5: Calculate the unreacted \( O_2 \)
Now, we can find out how much \( O_2 \) is left unreacted after the reaction:
\[ \text{Unreacted } O_2 = \text{Initial } O_2 - \text{Required } O_2 = 8.0 \text{ L} - 6.0 \text{ L} = 2.0 \text{ L} \]

### Conclusion
The amount of \( O_2 \) left unreacted is 2.0 L.

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