What volume of oxygen gas `(O_(2))` measured at `0^(@)C` and 1 atm, is needed to burn completely 1L of propane gas `(C_(3)H_(8))` measured under the same conditions ?

Options

A. 5 L
B. 10 L
C. 7 L
D. 6 L

Answer

To determine the volume of oxygen gas (O₂) required to completely burn 1 liter of propane gas (C₃H₈) at 0°C and 1 atm, we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of propane.
The combustion of propane can be represented by the following balanced equation:

\[ C_3H_8 + O_2 \rightarrow CO_2 + H_2O \]

### Step 2: Balance the equation.
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides. The balanced equation for the combustion of propane is:

\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \]

This indicates that 1 mole of propane reacts with 5 moles of oxygen.

### Step 3: Use the volume ratio.
According to Avogadro's law, at the same temperature and pressure, the volume of gases is directly proportional to the number of moles. Thus, we can use the volume ratio directly from the balanced equation.

From the balanced equation, we see that:
- 1 volume of C₃H₈ requires 5 volumes of O₂.

### Step 4: Calculate the volume of O₂ needed for 1L of C₃H₈.
Since we have 1 liter of propane (C₃H₈), we can calculate the volume of oxygen required:

\[ \text{Volume of } O_2 = 5 \times \text{Volume of } C_3H_8 = 5 \times 1 \text{ L} = 5 \text{ L} \]

### Conclusion:
Therefore, the volume of oxygen gas needed to completely burn 1 liter of propane gas at 0°C and 1 atm is **5 liters**.

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