If 50 ml of `A_(2)B_(3)` reacts completely with 200 ml of `C_(2)` in closed vessel according to the equation , `2A_(2)B_(3)(g)+5C_(2)(g) to 3C_(3)B_2(g)+CA_(4)(g)` . The composition of the mixture of gases is :

Options

A. 10 ml ` C_(2)` , 25 ml `C_(3)B_(2)`, 100 ml `CA_(4)`
B. 25 ml `C_(2)`, 75 ml ` C_(3),B_2,` 25 ml ` CA_(4)`
C. 75 ml ` C_(2)`, 75 ml ` C_(3)B_(2)`, 25 ml ` CA_(4)`
D. 100 ml `C_(2)`, 50 ml `C_(3)B_(2)`, 50 ml `CA_(4)`

Explanation

To solve the problem, we need to analyze the chemical reaction and the volumes of the gases involved. The reaction is given as:

\[ 2A_2B_3(g) + 5C_2(g) \rightarrow 3C_3B_2(g) + CA_4(g) \]

### Step 1: Identify the initial volumes of reactants
- Volume of \( A_2B_3 \) = 50 ml
- Volume of \( C_2 \) = 200 ml

### Step 2: Determine the stoichiometric ratios
From the balanced equation, the stoichiometric ratios are:
- 2 moles of \( A_2B_3 \) react with 5 moles of \( C_2 \).

### Step 3: Calculate the volume of \( C_2 \) required for the reaction
Since the ratio of \( A_2B_3 \) to \( C_2 \) is 2:5, we can set up the following proportion to find out how much \( C_2 \) is needed for the 50 ml of \( A_2B_3 \):

\[ \text{Volume of } C_2 \text{ required} = 50 \, \text{ml} \times \frac{5}{2} = 125 \, \text{ml} \]

### Step 4: Identify the limiting reagent
- We have 50 ml of \( A_2B_3 \) and need 125 ml of \( C_2 \) to completely react with it.
- We have 200 ml of \( C_2 \), which is more than enough.
- Therefore, \( A_2B_3 \) is the limiting reagent.

### Step 5: Calculate the remaining volume of \( C_2 \)
After the reaction, the remaining volume of \( C_2 \) can be calculated as follows:

\[ \text{Remaining } C_2 = \text{Initial } C_2 - \text{Used } C_2 = 200 \, \text{ml} - 125 \, \text{ml} = 75 \, \text{ml} \]

### Step 6: Calculate the volumes of the products formed
Using the stoichiometric ratios from the balanced equation:

1. Volume of \( C_3B_2 \) formed:
\[ \text{Volume of } C_3B_2 = 50 \, \text{ml} \times \frac{3}{2} = 75 \, \text{ml} \]

2. Volume of \( CA_4 \) formed:
\[ \text{Volume of } CA_4 = 50 \, \text{ml} \times \frac{1}{2} = 25 \, \text{ml} \]

### Step 7: Summarize the final composition of gases
After the reaction, the volumes of gases present in the vessel are:
- \( C_2 \): 75 ml
- \( C_3B_2 \): 75 ml
- \( CA_4 \): 25 ml

### Final Answer
The composition of the mixture of gases is:
- 75 ml of \( C_2 \)
- 75 ml of \( C_3B_2 \)
- 25 ml of \( CA_4 \)

Thus, the correct option is **Option C**.