10 ml of gaseous hydrocarbon on combution gives 20ml of `CO_(2)` and 30ml of `H_(2)O(g)`. The hydrocarbon is :-

Options

A. `C_(4)H_(5)`
B. `C_(2)H_(6)`
C. `C_(4)H_(8)`
D. `C_(4)H_(10)`

Explanation

To solve the problem of identifying the gaseous hydrocarbon from the given combustion data, we can follow these steps:

### Step 1: Write the general formula for the hydrocarbon
Assume the hydrocarbon is represented as \( C_xH_y \).

### Step 2: Write the balanced combustion reaction
The combustion of the hydrocarbon can be represented as:
\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

### Step 3: Determine the products formed
From the problem, we know:
- 10 mL of the hydrocarbon produces 20 mL of \( CO_2 \) and 30 mL of \( H_2O(g) \).

### Step 4: Relate the volumes of products to the coefficients in the balanced equation
From the balanced equation, we know:
- Each mole of \( C_xH_y \) produces \( x \) moles of \( CO_2 \) and \( \frac{y}{2} \) moles of \( H_2O \).

### Step 5: Set up the relationships based on the volumes
For the combustion of 10 mL of the hydrocarbon:
- The volume of \( CO_2 \) produced is 20 mL, which corresponds to \( x \) moles of \( CO_2 \).
- The volume of \( H_2O \) produced is 30 mL, which corresponds to \( \frac{y}{2} \) moles of \( H_2O \).

### Step 6: Calculate the values of \( x \) and \( y \)
From the volume of \( CO_2 \):
\[ \frac{20 \text{ mL}}{10 \text{ mL}} = 2 \Rightarrow x = 2 \]

From the volume of \( H_2O \):
\[ \frac{30 \text{ mL}}{10 \text{ mL}} = 3 \Rightarrow \frac{y}{2} = 3 \Rightarrow y = 6 \]

### Step 7: Write the molecular formula of the hydrocarbon
Now that we have \( x = 2 \) and \( y = 6 \), the molecular formula of the hydrocarbon is:
\[ C_2H_6 \]

### Conclusion
The hydrocarbon is ethane, represented by the formula \( C_2H_6 \).

---