When 10 ml of hydrogen and 12.5 ml of chlorine are allowed to react the final mixture contains under the same conditions

Options

A. 22.5 ml of HCl
B. 12.5 ml of HCl
C. 20 ml of HCl and 2.5 ml of chlorine
D. 20 ml of HCl only

Explanation

To solve the problem of the reaction between hydrogen and chlorine, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between hydrogen (H₂) and chlorine (Cl₂) produces hydrochloric acid (HCl). The balanced equation is:
\[ \text{H}_2 + \text{Cl}_2 \rightarrow 2 \text{HCl} \]

### Step 2: Determine the stoichiometry of the reaction
From the balanced equation, we can see that:
- 1 volume of H₂ reacts with 1 volume of Cl₂ to produce 2 volumes of HCl.

### Step 3: Calculate the initial volumes of reactants
We are given:
- Volume of H₂ = 10 mL
- Volume of Cl₂ = 12.5 mL

### Step 4: Identify the limiting reactant
According to the stoichiometry:
- 10 mL of H₂ would require 10 mL of Cl₂ to completely react (1:1 ratio).
- However, we have 12.5 mL of Cl₂ available.

Since we have enough Cl₂ to react with all of the H₂, H₂ is the limiting reactant.

### Step 5: Calculate the volume of HCl produced
From the balanced equation, 1 volume of H₂ produces 2 volumes of HCl. Therefore:
- 10 mL of H₂ will produce \( 10 \times 2 = 20 \) mL of HCl.

### Step 6: Calculate the remaining volume of Cl₂
Since 10 mL of Cl₂ is required to react with 10 mL of H₂, we can find the remaining Cl₂:
- Initial volume of Cl₂ = 12.5 mL
- Volume of Cl₂ used = 10 mL
- Remaining volume of Cl₂ = \( 12.5 - 10 = 2.5 \) mL

### Step 7: Summarize the final mixture
After the reaction, the final mixture will contain:
- 20 mL of HCl
- 2.5 mL of unreacted Cl₂

### Final Answer
The final mixture contains 20 mL of HCl and 2.5 mL of Cl₂.

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