How many litres of oxygen at 1 atm and 273 K will be required to burn completely 2.2 g of propane `(C_3H_8)` ?

Options

A. 11.2L
B. 22.4 L
C. 5.6 L
D. 44.8 L

Answer

To solve the problem of how many liters of oxygen at 1 atm and 273 K are required to completely burn 2.2 g of propane (C₃H₈), we will 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:
\[ C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O \]
This equation indicates that 1 mole of propane reacts with 5 moles of oxygen.

### Step 2: Calculate the number of moles of propane (C₃H₈) in 2.2 g.
To find the number of moles, we use the formula:
\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of propane (C₃H₈) is calculated as follows:
- Carbon (C): 12.01 g/mol × 3 = 36.03 g/mol
- Hydrogen (H): 1.008 g/mol × 8 = 8.064 g/mol
- Total molar mass of C₃H₈ = 36.03 g/mol + 8.064 g/mol = 44.094 g/mol (approximately 44 g/mol)

Now, calculate the moles of propane:
\[ \text{moles of } C_3H_8 = \frac{2.2 \, \text{g}}{44 \, \text{g/mol}} = 0.05 \, \text{moles} \]

### Step 3: Use the stoichiometry of the reaction to find the moles of oxygen (O₂) required.
From the balanced equation, we see that 1 mole of propane requires 5 moles of oxygen. Therefore:
\[ \text{moles of } O_2 = 5 \times \text{moles of } C_3H_8 = 5 \times 0.05 = 0.25 \, \text{moles} \]

### Step 4: Calculate the volume of oxygen gas at STP (Standard Temperature and Pressure).
At STP (273 K and 1 atm), 1 mole of any ideal gas occupies 22.4 liters. Thus, the volume of oxygen required can be calculated as:
\[ \text{Volume of } O_2 = \text{moles of } O_2 \times 22.4 \, \text{L/mol} \]
\[ \text{Volume of } O_2 = 0.25 \, \text{moles} \times 22.4 \, \text{L/mol} = 5.6 \, \text{L} \]

### Final Answer:
The volume of oxygen required to burn 2.2 g of propane completely is **5.6 liters**.

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