A sample of `NH_(3)` gas is 20% dissociated into `N_(2)` and `H_(2)` gases. The mass ratio of `N_(2)` and `NH_(3)` gases in the final sample is

Options

A. `(7)/(34)`
B. `(7)/(17)`
C. `(14)/(17)`
D. `(21)/(17)`

Answer

To solve the problem, we need to determine the mass ratio of \( N_2 \) to \( NH_3 \) after 20% of \( NH_3 \) has dissociated into \( N_2 \) and \( H_2 \).

### Step-by-Step Solution:

1. **Initial Setup:**
- Assume we start with 1 mole of \( NH_3 \).

2. **Dissociation Calculation:**
- Given that 20% of \( NH_3 \) dissociates:
\[ \text{Dissociated moles of } NH_3 = 0.20 \times 1 = 0.2 \text{ moles} \]

3. **Stoichiometry of Dissociation:**
- The dissociation of \( NH_3 \) can be represented as:
\[ 2 NH_3 \rightarrow N_2 + 3 H_2 \]
- From the stoichiometry, 2 moles of \( NH_3 \) produce 1 mole of \( N_2 \) and 3 moles of \( H_2 \).
- Therefore, from 0.2 moles of \( NH_3 \):
- Moles of \( N_2 \) produced:
\[ \text{Moles of } N_2 = \frac{0.2}{2} = 0.1 \text{ moles} \]
- Moles of \( H_2 \) produced:
\[ \text{Moles of } H_2 = \frac{3}{2} \times 0.2 = 0.3 \text{ moles} \]

4. **Remaining Moles of \( NH_3 \):**
- Moles of \( NH_3 \) remaining after dissociation:
\[ \text{Remaining } NH_3 = 1 - 0.2 = 0.8 \text{ moles} \]

5. **Mass Calculation:**
- Molar mass of \( N_2 = 28 \, g/mol \)
- Molar mass of \( NH_3 = 17 \, g/mol \)
- Mass of \( N_2 \) produced:
\[ \text{Mass of } N_2 = 0.1 \text{ moles} \times 28 \, g/mol = 2.8 \, g \]
- Mass of remaining \( NH_3 \):
\[ \text{Mass of } NH_3 = 0.8 \text{ moles} \times 17 \, g/mol = 13.6 \, g \]

6. **Mass Ratio Calculation:**
- The mass ratio of \( N_2 \) to \( NH_3 \):
\[ \text{Mass ratio} = \frac{\text{Mass of } N_2}{\text{Mass of } NH_3} = \frac{2.8 \, g}{13.6 \, g} = \frac{2.8}{13.6} = \frac{7}{34} \]

### Final Answer:
The mass ratio of \( N_2 \) to \( NH_3 \) in the final sample is \( \frac{7}{34} \).