For the gaseous reaction `H_(2(g)+Cl_(2)(g)to2HCl(g)` If initially 40ml of `H_(2)` and 30 ml of `Cl_(2)` are present the find the volume of unreacted part of `Cl_(2)(g)`:-

Options

A. 10 ml
B. 25 ml
C. 15 ml
D. 0 ml

Answer

To solve the problem, we need to analyze the given reaction and the initial volumes of the reactants. The reaction is:

\[ H_2(g) + Cl_2(g) \rightarrow 2HCl(g) \]

### Step 1: Identify the initial volumes of the reactants
- Initial volume of \( H_2 \) = 40 ml
- Initial volume of \( Cl_2 \) = 30 ml

### Step 2: Determine the stoichiometry of the reaction
From the balanced equation, we see that:
- 1 volume of \( H_2 \) reacts with 1 volume of \( Cl_2 \) to produce 2 volumes of \( HCl \).

### Step 3: Identify the limiting reagent
To find out which reactant is the limiting reagent, we compare the initial volumes:
- We have 40 ml of \( H_2 \) and 30 ml of \( Cl_2 \).
- According to the stoichiometry, 30 ml of \( Cl_2 \) would require 30 ml of \( H_2 \) to completely react.

### Step 4: Calculate the amount of reactants consumed
Since \( Cl_2 \) is the limiting reagent, it will be completely consumed:
- Volume of \( Cl_2 \) consumed = 30 ml
- Volume of \( H_2 \) consumed = 30 ml (since it reacts in a 1:1 ratio with \( Cl_2 \))

### Step 5: Calculate the remaining volume of each reactant
- Remaining volume of \( H_2 \) = Initial volume of \( H_2 \) - Volume of \( H_2 \) consumed
\[ = 40 \, \text{ml} - 30 \, \text{ml} = 10 \, \text{ml} \]
- Remaining volume of \( Cl_2 \) = Initial volume of \( Cl_2 \) - Volume of \( Cl_2 \) consumed
\[ = 30 \, \text{ml} - 30 \, \text{ml} = 0 \, \text{ml} \]

### Conclusion
The volume of unreacted \( Cl_2 \) is 0 ml.

### Final Answer
The volume of unreacted \( Cl_2(g) \) is **0 ml**.

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