Ten millilitre of a gaseous hydrocarbon was burnt completely in 80 ml of `O_2` at STP. The volume of the remaining gas is 70 ml. The volume became 50 ml, on treatment with `NaOH`. The formula of the hydrocarbon is:

Options

A. `C_2H_6`
B. `C_2H_4`
C. `C_3H_8`
D. `C_3H_6`

Explanation

To solve the problem step by step, we will analyze the combustion of the gaseous hydrocarbon and the subsequent reactions.

### Step 1: Identify the hydrocarbon
Let the hydrocarbon be represented as \( C_xH_y \).

### Step 2: Write the balanced combustion reaction
When the hydrocarbon burns in oxygen, it produces carbon dioxide and water:
\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

### Step 3: Determine the volumes of gases involved
From the problem, we know:
- Volume of hydrocarbon burnt = 10 mL
- Volume of oxygen used = 80 mL
- Volume of remaining gas after combustion = 70 mL

### Step 4: Calculate the volume of unreacted oxygen
The volume of gas remaining after combustion is 70 mL. This volume consists of unreacted oxygen and possibly other gases. The volume of unreacted oxygen can be calculated as:
\[ \text{Remaining gas} = \text{Unreacted O}_2 + \text{CO}_2 + \text{H}_2O \]
Since water is in vapor form and not counted in the gas volume, we can focus on the unreacted oxygen and carbon dioxide.

### Step 5: Calculate the volume of carbon dioxide produced
When the gas is treated with NaOH, the volume reduces to 50 mL. NaOH absorbs carbon dioxide, so the change in volume (70 mL to 50 mL) indicates the volume of CO2 produced:
\[ \text{Volume of CO}_2 = 70 \text{ mL} - 50 \text{ mL} = 20 \text{ mL} \]

### Step 6: Relate the volume of CO2 to the hydrocarbon
From the combustion reaction, we know that:
\[ \text{Volume of CO}_2 = 10x \text{ mL} \]
Setting this equal to the volume of CO2 produced:
\[ 10x = 20 \implies x = 2 \]

### Step 7: Determine the volume of unreacted oxygen
Now, we can find the volume of unreacted oxygen. The total volume of oxygen initially was 80 mL. The volume of oxygen consumed can be calculated using the stoichiometry of the reaction:
\[ \text{Volume of O}_2 = \frac{x + \frac{y}{4}}{10} \text{ mL} \]
Substituting \( x = 2 \):
\[ \text{Volume of O}_2 = 80 - \left(2 + \frac{y}{4}\right) \cdot 10 \]
The remaining volume after combustion is 70 mL, which includes unreacted oxygen:
\[ 70 = 80 - (2 + \frac{y}{4}) \cdot 10 \]
Solving for \( y \):
\[ 70 = 80 - 20 - 2y \implies 2y = 10 \implies y = 5 \]

### Step 8: Write the final formula of the hydrocarbon
Now we have \( x = 2 \) and \( y = 5 \), so the formula of the hydrocarbon is:
\[ C_2H_5 \]

### Conclusion
The formula of the hydrocarbon is \( C_2H_5 \), which corresponds to ethane.