`40%` of a mixture of `0.2` 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 step by step, we will follow the provided information and apply the principles of chemical equilibrium and stoichiometry.

### Step 1: Identify the Initial Moles
We have a mixture of:
- \( N_2 \): 0.2 moles
- \( H_2 \): 0.6 moles

### Step 2: Calculate the Total Initial Moles
The total initial moles of gases is:
\[ \text{Total Initial Moles} = 0.2 + 0.6 = 0.8 \text{ moles} \]

### Step 3: Determine the Amount of Reactants that React
According to the question, 40% of the mixture reacts. Therefore, the amount of the mixture that reacts is:
\[ \text{Moles that react} = 0.4 \times 0.8 = 0.32 \text{ moles} \]

### Step 4: Set Up the Reaction Stoichiometry
The balanced chemical equation is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]
From the stoichiometry of the reaction:
- 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \) to produce 2 moles of \( NH_3 \).

### Step 5: Define Variables for the Reaction
Let \( x \) be the moles of \( N_2 \) that react. Then:
- Moles of \( H_2 \) that react = \( 3x \)
- Moles of \( NH_3 \) produced = \( 2x \)

### Step 6: Set Up the Equations Based on Reactants
From the information provided, we know that:
\[ x + 3x = 0.32 \text{ (total moles that react)} \]
This simplifies to:
\[ 4x = 0.32 \implies x = 0.08 \]

### Step 7: Calculate Final Moles of Each Component
Now we can calculate the final moles of each component:
- Moles of \( N_2 \) remaining = \( 0.2 - x = 0.2 - 0.08 = 0.12 \)
- Moles of \( H_2 \) remaining = \( 0.6 - 3x = 0.6 - 3(0.08) = 0.6 - 0.24 = 0.36 \)
- Moles of \( NH_3 \) produced = \( 2x = 2(0.08) = 0.16 \)

### Step 8: Calculate Total Final Moles
The total final moles of gases is:
\[ \text{Total Final Moles} = \text{Moles of } N_2 + \text{Moles of } H_2 + \text{Moles of } NH_3 \]
\[ = 0.12 + 0.36 + 0.16 = 0.64 \text{ moles} \]

### Step 9: Calculate the Ratio of Final Volume to Initial Volume
The ratio of final volume to initial volume is given by the ratio of final moles to initial moles:
\[ \text{Ratio} = \frac{\text{Final Moles}}{\text{Initial Moles}} = \frac{0.64}{0.8} \]
\[ = \frac{64}{80} = \frac{4}{5} \]

### Final Answer
The ratio of the final volume to the initial volume of gases is \( 4:5 \).

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