If `1:3` (molar ratio) mixture of `N_2` and `H_2` yields 20% (by volume) of `NH_3` at `30 atm`, then the moles of `N_2` converted into the product at equilibrium will be

Options

A. `0.66`
B. `0.22`
C. `0.33`
D. `0.44`

Explanation

To solve the problem, we need to determine how many moles of \( N_2 \) are converted into \( NH_3 \) at equilibrium when starting with a \( 1:3 \) molar ratio of \( N_2 \) and \( H_2 \) and yielding \( 20\% \) by volume of \( NH_3 \) at \( 30 \, atm \).

### Step-by-Step Solution:

1. **Identify Initial Moles**:
- Assume we start with \( 1 \) mole of \( N_2 \).
- Since the ratio is \( 1:3 \), we have \( 3 \) moles of \( H_2 \).
- Initial moles: \( N_2 = 1 \), \( H_2 = 3 \), \( NH_3 = 0 \).

2. **Define Change in Moles**:
- Let \( X \) be the moles of \( N_2 \) that react.
- The reaction is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]
- Therefore, if \( X \) moles of \( N_2 \) react, then:
- Moles of \( N_2 \) left = \( 1 - X \)
- Moles of \( H_2 \) left = \( 3 - 3X \)
- Moles of \( NH_3 \) produced = \( 2X \)

3. **Calculate Total Moles at Equilibrium**:
- Total moles at equilibrium = Moles of \( N_2 \) + Moles of \( H_2 \) + Moles of \( NH_3 \)
- Total moles = \( (1 - X) + (3 - 3X) + 2X = 4 - 2X \)

4. **Relate Volume and Moles**:
- We know that the volume percentage of \( NH_3 \) is \( 20\% \).
- Since the volume is proportional to the number of moles, we can set up the equation:
\[ \frac{2X}{4 - 2X} = \frac{20}{100} = \frac{1}{5} \]

5. **Cross-Multiply and Solve for \( X \)**:
- Cross-multiplying gives:
\[ 2X \cdot 5 = 1 \cdot (4 - 2X) \]
\[ 10X = 4 - 2X \]
\[ 10X + 2X = 4 \]
\[ 12X = 4 \]
\[ X = \frac{4}{12} = \frac{1}{3} \approx 0.33 \]

6. **Conclusion**:
- The moles of \( N_2 \) converted into \( NH_3 \) at equilibrium is \( \frac{1}{3} \) moles or approximately \( 0.33 \) moles.