How many litres of oxygen at 1 atm & and 273K will be required to burn completely 2.2 of propane `(C_(3)H_(8))`

Options

A. 11.2 L
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 burn completely 2.2 grams of propane (C₃H₈), we can follow these steps:

### Step 1: Calculate the Molar Mass of Propane (C₃H₈)
The molar mass of propane is calculated as follows:
- Carbon (C) has a molar mass of approximately 12 g/mol.
- Hydrogen (H) has a molar mass of approximately 1 g/mol.

For propane (C₃H₈):
\[ \text{Molar mass of C₃H₈} = (3 \times 12) + (8 \times 1) = 36 + 8 = 44 \text{ g/mol} \]

### Step 2: Calculate the Moles of Propane
Using the formula:
\[ \text{Moles of propane} = \frac{\text{mass of propane}}{\text{molar mass of propane}} \]
Substituting the values:
\[ \text{Moles of propane} = \frac{2.2 \text{ g}}{44 \text{ g/mol}} = 0.05 \text{ moles} \]

### Step 3: Write the Balanced Equation for the Combustion of Propane
The balanced chemical equation for the combustion of propane is:
\[ \text{C₃H₈} + 5 \text{O₂} \rightarrow 3 \text{CO₂} + 4 \text{H₂O} \]
This indicates that 1 mole of propane reacts with 5 moles of oxygen.

### Step 4: Calculate the Moles of Oxygen Required
From the balanced equation, we see that:
\[ 1 \text{ mole of C₃H₈} \text{ requires } 5 \text{ moles of O₂} \]
Thus, for 0.05 moles of propane:
\[ \text{Moles of O₂ required} = 5 \times 0.05 = 0.25 \text{ moles} \]

### Step 5: Convert Moles of Oxygen to Liters at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of oxygen required can be calculated as:
\[ \text{Volume of O₂} = \text{moles of O₂} \times 22.4 \text{ L/mol} \]
Substituting the values:
\[ \text{Volume of O₂} = 0.25 \text{ moles} \times 22.4 \text{ L/mol} = 5.6 \text{ liters} \]

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

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