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

Options

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

Explanation

To determine the volume of oxygen gas (O₂) needed to completely burn 1 liter of propane gas (C₃H₈) at standard temperature and pressure (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 equation:

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

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

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

### Step 3: Analyze the stoichiometry of the reaction.
From the balanced equation, we can see that:

- 1 mole of propane (C₃H₈) reacts with 5 moles of oxygen (O₂).

### Step 4: Relate the volumes of gases using the ideal gas law.
At standard temperature and pressure (0°C and 1 atm), the volumes of gases are directly proportional to the number of moles. Therefore, we can say:

- 1 liter of propane will require 5 liters of oxygen for complete combustion.

### Step 5: Calculate the volume of oxygen needed.
Since we have 1 liter of propane, the volume of oxygen required will be:

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

### Conclusion:
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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