`40%` of a mixture of `2.0` mol of `N_(2)` and `0.6` mol of `H_(2)` reacts to give `NH_(3)` according to the equation:
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`
at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

Options

A. `4:5`
B. `5:4`
C. `7:10`
D. `8:5`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the initial moles of reactants
We have a mixture of:
- Nitrogen (N₂): 2.0 mol
- Hydrogen (H₂): 0.6 mol

### Step 2: Calculate the total initial moles
Total initial moles = moles of N₂ + moles of H₂ = 2.0 mol + 0.6 mol = 2.6 mol

### Step 3: Calculate the amount of the mixture that reacts
40% of the mixture reacts. Therefore, the moles that react can be calculated as:
- Moles that react = 40% of total initial moles = 0.4 × 2.6 mol = 1.04 mol

### Step 4: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 5: Set up the stoichiometry of the reaction
From the balanced equation, we see that:
- 1 mol of N₂ reacts with 3 mol of H₂ to produce 2 mol of NH₃.

### Step 6: Determine the moles of N₂ and H₂ that react
Let \( x \) be the moles of N₂ that react. Then:
- Moles of N₂ reacted = \( x \)
- Moles of H₂ reacted = \( 3x \)

### Step 7: Relate the reacted moles to the total moles that reacted
From the previous steps, we know that:
\[ x + 3x = 1.04 \]
\[ 4x = 1.04 \]
\[ x = 0.26 \]

### Step 8: Calculate the moles of each substance at equilibrium
- Moles of N₂ remaining = Initial moles of N₂ - Moles of N₂ reacted = \( 2.0 - 0.26 = 1.74 \) mol
- Moles of H₂ remaining = Initial moles of H₂ - Moles of H₂ reacted = \( 0.6 - 3(0.26) = 0.6 - 0.78 = -0.18 \) mol (This indicates that H₂ is the limiting reactant)
- Moles of NH₃ produced = \( 2 \times 0.26 = 0.52 \) mol

### Step 9: Calculate the final moles of gases
Since H₂ is the limiting reactant, we will have:
- Moles of N₂ = 1.74 mol
- Moles of H₂ = 0 (all reacted)
- Moles of NH₃ = 0.52 mol

### Step 10: Calculate the final volume of gases
Final volume of gases = Moles of N₂ + Moles of NH₃ = \( 1.74 + 0.52 = 2.26 \) mol

### Step 11: Calculate the initial volume of gases
Initial volume of gases = Moles of N₂ + Moles of H₂ = \( 2.0 + 0.6 = 2.6 \) mol

### Step 12: Calculate the ratio of final volume to initial volume
Ratio = Final volume / Initial volume = \( \frac{2.26}{2.6} \)

### Step 13: Simplify the ratio
To simplify:
\[ \frac{2.26}{2.6} = \frac{226}{260} = \frac{113}{130} \]

### Conclusion
The ratio of the final volume to the initial volume of gases is \( \frac{113}{130} \).

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