The volume of oxygen necessary for the complete combustion of 20 litre of propane is :

Options

A. 40 litre
B. 60 litre
C. 80 litre
D. 100 litre

Answer

To find the volume of oxygen necessary for the complete combustion of 20 liters of propane (C₃H₈), we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of propane.
The balanced equation for the combustion of propane is:
\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \]

### Step 2: Identify the stoichiometric ratio between propane and oxygen.
From the balanced equation, we can see that 1 mole of propane reacts with 5 moles of oxygen. This means the ratio of propane to oxygen is:
\[ \text{Propane} : \text{Oxygen} = 1 : 5 \]

### Step 3: Calculate the volume of oxygen needed for 20 liters of propane.
Since the volume of gases at the same temperature and pressure is directly proportional to the number of moles, we can use the stoichiometric ratio to find the volume of oxygen required.

Given that we have 20 liters of propane, we can set up the proportion:
\[ \text{Volume of } O_2 = 5 \times \text{Volume of } C_3H_8 \]
\[ \text{Volume of } O_2 = 5 \times 20 \, \text{liters} = 100 \, \text{liters} \]

### Step 4: Conclusion
The volume of oxygen necessary for the complete combustion of 20 liters of propane is:
\[ \text{Volume of } O_2 = 100 \, \text{liters} \]

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