12 litre of `H_(2)` and 11.2 litre of `Cl_(2)` are mixed and exploded. The composition by volume of mixture is-

Options

A. 24 litre of HCl
B. 0.8 litre `Cl_(2)` and 20.8 lit HCl.
C. 0.8 litre `H_(2)` & 22.4 litre HCl
D. 22.4 litre HCl

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Write the balanced chemical equation
The first step is to write the balanced chemical equation for the reaction between hydrogen (H₂) and chlorine (Cl₂):
\[ \text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl} \]

### Step 2: Identify the stoichiometric coefficients
From the balanced equation, we can see that:
- 1 mole of H₂ reacts with 1 mole of Cl₂ to produce 2 moles of HCl.

### Step 3: Convert the volumes of gases to moles
We know that at standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.

For H₂:
\[ \text{Moles of H}_2 = \frac{12 \text{ liters}}{22.4 \text{ liters/mole}} = 0.5357 \text{ moles} \]

For Cl₂:
\[ \text{Moles of Cl}_2 = \frac{11.2 \text{ liters}}{22.4 \text{ liters/mole}} = 0.50 \text{ moles} \]

### Step 4: Determine the limiting reactant
From the stoichiometry of the reaction, we know that 1 mole of H₂ reacts with 1 mole of Cl₂.

- We have 0.5357 moles of H₂ and 0.50 moles of Cl₂.
- Since we need equal moles of H₂ and Cl₂ to react completely, Cl₂ is the limiting reactant because we have less of it.

### Step 5: Calculate the amount of H₂ used and remaining
Since Cl₂ is the limiting reactant, it will determine how much H₂ is used:
- Moles of H₂ used = Moles of Cl₂ = 0.50 moles

Now, calculate the remaining moles of H₂:
\[ \text{Remaining moles of H}_2 = 0.5357 - 0.50 = 0.0357 \text{ moles} \]

### Step 6: Convert remaining moles of H₂ to volume
To find the volume of the remaining H₂:
\[ \text{Volume of remaining H}_2 = 0.0357 \text{ moles} \times 22.4 \text{ liters/mole} = 0.799 \text{ liters} \approx 0.8 \text{ liters} \]

### Step 7: Calculate the amount of HCl produced
From the balanced equation, we know that 1 mole of Cl₂ produces 2 moles of HCl:
\[ \text{Moles of HCl produced} = 2 \times 0.50 = 1 \text{ mole} \]

Now, convert the moles of HCl to volume:
\[ \text{Volume of HCl} = 1 \text{ mole} \times 22.4 \text{ liters/mole} = 22.4 \text{ liters} \]

### Step 8: Summarize the composition by volume of the mixture
After the reaction, we have:
- Remaining H₂: 0.8 liters
- Produced HCl: 22.4 liters

### Final Answer
The composition by volume of the mixture is:
- H₂: 0.8 liters
- HCl: 22.4 liters