500 ml of a gaseous hydrocarbon when burnt in excess of `O_(2)` gave 2.0litres of `CO_(2)` and 2.5 litres of water vapours under same
conditions. molecular formula of the hydrocarbon is:-

Options

A. `C_(4)H_(8)`
B. `C_(4)H_(10)`
C. `C_(5)H_(10)`
D. `C_(5)H_(12)`

Explanation

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

### Step 1: Understand the combustion reaction
The combustion of a hydrocarbon (CxHy) in the presence of oxygen (O2) produces carbon dioxide (CO2) and water (H2O). The general reaction can be written as:
\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

### Step 2: Write the balanced equation
For the combustion of a hydrocarbon, the balanced equation can be represented as:
\[ C_xH_y + \frac{x + \frac{y}{4}}{O_2} \rightarrow x CO_2 + \frac{y}{2} H_2O \]

### Step 3: Determine the volume ratios
From the problem, we know:
- 500 ml of hydrocarbon produces 2.0 liters of CO2 and 2.5 liters of water vapor.
- Convert these volumes to milliliters:
- 2.0 liters = 2000 ml
- 2.5 liters = 2500 ml

### Step 4: Set up equations based on the volumes
From the combustion reaction:
- The volume of CO2 produced is proportional to the number of moles of carbon (x):
\[ 500x = 2000 \]
- The volume of water produced is proportional to half the number of moles of hydrogen (y):
\[ 250y = 2500 \]

### Step 5: Solve for x
From the first equation:
\[ 500x = 2000 \]
\[ x = \frac{2000}{500} = 4 \]

### Step 6: Solve for y
From the second equation:
\[ 250y = 2500 \]
\[ y = \frac{2500}{250} = 10 \]

### Step 7: Write the molecular formula
Now that we have the values of x and y:
- \( x = 4 \)
- \( y = 10 \)

The molecular formula of the hydrocarbon is:
\[ C_4H_{10} \]

### Final Answer
The molecular formula of the hydrocarbon is \( C_4H_{10} \).

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