500 ml of a hydrocarbon gas burnt in excess of oxygen yields 2500 ml of `CO_2` and 3 litres of water vapours. All volume being measured at the same temperature and pressure. The formula of the hydrocarbon is :

Options

A. `C_5H_10`
B. `C_5H_12`
C. `C_4H_6`
D. `C_3H_6`

Answer

To find the formula of the hydrocarbon, we can follow these steps:

### Step 1: Understand the Reaction
We start with a hydrocarbon (let's assume its formula is \( C_xH_y \)) that burns in excess oxygen to produce carbon dioxide (\( CO_2 \)) and water vapor (\( H_2O \)). The reaction can be represented as:
\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

### Step 2: Write the Balanced Equation
From the combustion of the hydrocarbon, we can write the balanced equation:
\[ C_xH_y + \left( \frac{x + y/4}{2} \right) O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O \]

### Step 3: Use Volume Relationships
Given that 500 ml of hydrocarbon produces 2500 ml of \( CO_2 \) and 3 liters (3000 ml) of water vapor, we can use the fact that at constant temperature and pressure, the volume of gases is proportional to the number of moles.

From the problem:
- Volume of hydrocarbon = 500 ml
- Volume of \( CO_2 \) produced = 2500 ml
- Volume of \( H_2O \) produced = 3000 ml

### Step 4: Relate Volumes to Coefficients
From the balanced equation:
- For every 1 volume of hydrocarbon, \( x \) volumes of \( CO_2 \) and \( \frac{y}{2} \) volumes of \( H_2O \) are produced.

Thus, we can set up the equations:
1. \( 500 \cdot x = 2500 \) (for \( CO_2 \))
2. \( 500 \cdot \frac{y}{2} = 3000 \) (for \( H_2O \))

### Step 5: Solve for \( x \)
From the first equation:
\[ 500x = 2500 \]
\[ x = \frac{2500}{500} = 5 \]

### Step 6: Solve for \( y \)
From the second equation:
\[ 500 \cdot \frac{y}{2} = 3000 \]
\[ \frac{y}{2} = \frac{3000}{500} \]
\[ \frac{y}{2} = 6 \]
\[ y = 12 \]

### Step 7: Write the Formula of the Hydrocarbon
Now that we have \( x = 5 \) and \( y = 12 \), the formula of the hydrocarbon is:
\[ C_5H_{12} \]

### Conclusion
The formula of the hydrocarbon is \( C_5H_{12} \).

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