How many litre of oxygen at STP is required to burn 60 g `C_(2)H_(6)`?

Options

A. 22.4L
B. 11.21
C. `22.4 xx 7L`
D. 8.5 L

Answer

To determine how many liters of oxygen at STP are required to burn 60 g of ethane (C₂H₆), we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of ethane.
The combustion of ethane can be represented as:
\[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]
Balancing this equation, we get:
\[ \text{C}_2\text{H}_6 + \frac{7}{2} \text{O}_2 \rightarrow 2 \text{CO}_2 + 3 \text{H}_2\text{O} \]
This can also be written as:
\[ 2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2\text{O} \]

### Step 2: Calculate the molar mass of ethane (C₂H₆).
The molar mass of C₂H₆ can be calculated as follows:
- Carbon (C): 12 g/mol × 2 = 24 g/mol
- Hydrogen (H): 1 g/mol × 6 = 6 g/mol
- Total molar mass of C₂H₆ = 24 g/mol + 6 g/mol = 30 g/mol

### Step 3: Determine the moles of ethane in 60 g.
Using the molar mass calculated:
\[ \text{Moles of C}_2\text{H}_6 = \frac{\text{mass}}{\text{molar mass}} = \frac{60 \, \text{g}}{30 \, \text{g/mol}} = 2 \, \text{moles} \]

### Step 4: Use the balanced equation to find the moles of oxygen required.
From the balanced equation, we see that 1 mole of C₂H₆ requires 3.5 moles of O₂. Therefore, for 2 moles of C₂H₆:
\[ \text{Moles of O}_2 = 2 \, \text{moles C}_2\text{H}_6 \times \frac{7}{2} \, \text{moles O}_2/\text{mole C}_2\text{H}_6 = 7 \, \text{moles O}_2 \]

### Step 5: Calculate the volume of oxygen at STP.
At STP, 1 mole of any gas occupies 22.4 liters. Therefore, the volume of 7 moles of O₂ is:
\[ \text{Volume of O}_2 = 7 \, \text{moles} \times 22.4 \, \text{L/mole} = 156.8 \, \text{L} \]

### Final Answer:
The volume of oxygen required to burn 60 g of ethane at STP is **156.8 liters**.

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