Mixture of 1 mole of each `CH_(4)` and `C_(2)H_(6)` absorb 5 mole of `Cl_(2)` to form `C Cl_(4)` and `C_(2)Cl_(6)`. Then average molecular wt. of the gaseous mixture is

Options

A. 20
B. 23
C. 35
D. 15

Answer

To find the average molecular weight of the gaseous mixture after the reaction, we can follow these steps:

### Step 1: Identify the reactants and products
We have a mixture of 1 mole of methane (CH₄) and 1 mole of ethane (C₂H₆) that reacts with chlorine (Cl₂) to form carbon tetrachloride (CCl₄) and dichloroethane (C₂Cl₆).

### Step 2: Write the balanced chemical equation
The balanced reaction can be represented as:
\[ \text{CH}_4 + \text{C}_2\text{H}_6 + 5 \text{Cl}_2 \rightarrow \text{CCl}_4 + \text{C}_2\text{Cl}_6 + 10 \text{HCl} \]

### Step 3: Determine the amount of Cl₂ absorbed
According to the problem, 5 moles of Cl₂ are absorbed. However, from the stoichiometry of the balanced equation, we see that 1 mole of CH₄ and 1 mole of C₂H₆ would require 10 moles of Cl₂. Since only 5 moles are available, this means that only part of each hydrocarbon will react.

### Step 4: Calculate the moles of CH₄ and C₂H₆ that react
Since 5 moles of Cl₂ are absorbed, we can find out how much of CH₄ and C₂H₆ are used:
- The ratio of Cl₂ to CH₄ and C₂H₆ is 10:1 (for 1 mole of each hydrocarbon).
- Therefore, if 5 moles of Cl₂ are used, then:
\[ \text{Moles of CH}_4 \text{ used} = \frac{5 \text{ moles Cl}_2}{10} = 0.5 \text{ moles} \]
\[ \text{Moles of C}_2\text{H}_6 \text{ used} = \frac{5 \text{ moles Cl}_2}{10} = 0.5 \text{ moles} \]

### Step 5: Calculate the molecular weights
Now we calculate the molecular weights of the hydrocarbons:
- Molecular weight of CH₄ = 12 (C) + 4 (H) = 16 g/mol
- Molecular weight of C₂H₆ = 2(12) + 6(1) = 30 g/mol

### Step 6: Calculate the average molecular weight of the mixture
Now we can calculate the average molecular weight of the gaseous mixture:
\[ \text{Average Molecular Weight} = \frac{(0.5 \text{ moles} \times 16 \text{ g/mol}) + (0.5 \text{ moles} \times 30 \text{ g/mol})}{0.5 + 0.5} \]
\[ = \frac{(8 + 15)}{1} = 23 \text{ g/mol} \]

### Conclusion
The average molecular weight of the gaseous mixture after the reaction is **23 g/mol**.