`5L` of an alkane requires `25L` of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is :

Options

A. Butane
B. Isobutane
C. Propane
D. Ethane

Answer

To determine which alkane corresponds to the given combustion data, we can follow these steps:

### Step 1: Understand the combustion reaction
The general combustion reaction for an alkane (CnH2n+2) can be represented as:
\[ \text{CnH}_{2n+2} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

### Step 2: Write the balanced equation
For the complete combustion of an alkane, the balanced equation can be derived. The number of moles of oxygen required can be calculated using the formula:
\[ \text{Oxygen required} = \frac{3n + 1}{2} \]

### Step 3: Set up the volume relationship
Given that 5L of alkane requires 25L of oxygen, we can establish a relationship between the volumes and moles:
- Let \( V_1 = 5 \, \text{L} \) (volume of alkane)
- Let \( V_2 = 25 \, \text{L} \) (volume of oxygen)

Using the ideal gas law, we know that at constant temperature and pressure, the volume is directly proportional to the number of moles:
\[ \frac{V_2}{V_1} = \frac{n_2}{n_1} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ \frac{25}{5} = \frac{n_2}{1} \]
This simplifies to:
\[ 5 = n_2 \]

### Step 5: Relate \( n_2 \) to the alkane
From our earlier equation for the combustion of the alkane, we have:
\[ n_2 = \frac{3n + 1}{2} \]
Setting \( n_2 = 5 \):
\[ 5 = \frac{3n + 1}{2} \]

### Step 6: Solve for \( n \)
To solve for \( n \), multiply both sides by 2:
\[ 10 = 3n + 1 \]
Subtract 1 from both sides:
\[ 9 = 3n \]
Now, divide by 3:
\[ n = 3 \]

### Step 7: Identify the alkane
Using the value of \( n \):
- For \( n = 1 \): Methane (C1H4)
- For \( n = 2 \): Ethane (C2H6)
- For \( n = 3 \): Propane (C3H8)

Thus, the alkane is **Propane (C3H8)**.

### Final Answer:
The alkane is **Propane (C3H8)**.

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