A cylinder contains either ethylene or propylene 12 ml of gas required 54 ml of oxygen for complete combustion. The gas is

Options

A. Ethylene
B. Propylene
C. `1 : 1` mixture of two gases
D. `1 : 2 `mixture

Answer

To determine whether the gas in the cylinder is ethylene (C2H4) or propylene (C3H6), we need to analyze the stoichiometry of the combustion reactions for both gases and compare the amount of oxygen required for complete combustion.

### Step-by-Step Solution:

1. **Write the Combustion Reactions**:
- For ethylene (C2H4):
\[ C_2H_4 + O_2 \rightarrow CO_2 + H_2O \]
- For propylene (C3H6):
\[ C_3H_6 + O_2 \rightarrow CO_2 + H_2O \]

2. **Balance the Combustion Reaction for Ethylene**:
- The balanced equation for ethylene is:
\[ C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O \]
- This indicates that 1 mole of ethylene requires 3 moles of oxygen.

3. **Calculate the Oxygen Required for 12 ml of Ethylene**:
- If 12 ml of ethylene is used, the oxygen required would be:
\[ 3 \times 12 \text{ ml} = 36 \text{ ml} \]

4. **Balance the Combustion Reaction for Propylene**:
- The balanced equation for propylene is:
\[ C_3H_6 + \frac{9}{2} O_2 \rightarrow 3CO_2 + 3H_2O \]
- This indicates that 1 mole of propylene requires 4.5 moles of oxygen.

5. **Calculate the Oxygen Required for 12 ml of Propylene**:
- If 12 ml of propylene is used, the oxygen required would be:
\[ \frac{9}{2} \times 12 \text{ ml} = 54 \text{ ml} \]

6. **Compare the Oxygen Requirements**:
- The problem states that 54 ml of oxygen is required for complete combustion.
- Since 54 ml of oxygen corresponds to the combustion of 12 ml of propylene, we conclude that the gas in the cylinder is propylene.

### Final Answer:
The gas in the cylinder is **propylene (C3H6)**.