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 \( C_2H_6 \) (ethane), we can follow these steps:

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

\[ C_2H_6 + \frac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O \]

### Step 2: Calculate the molar mass of \( C_2H_6 \).
The molar mass of ethane (\( C_2H_6 \)) 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 \( C_2H_6 \) = 24 g/mol + 6 g/mol = 30 g/mol

### Step 3: Determine the number of moles of \( C_2H_6 \) in 60 g.
Using the molar mass, we can find the number of moles of \( C_2H_6 \):

\[ \text{Number of moles of } C_2H_6 = \frac{\text{mass}}{\text{molar mass}} = \frac{60 \text{ g}}{30 \text{ g/mol}} = 2 \text{ moles} \]

### Step 4: Use the stoichiometry of the reaction to find the moles of \( O_2 \) required.
From the balanced equation, we see that 1 mole of \( C_2H_6 \) requires \( \frac{7}{2} \) moles of \( O_2 \). Therefore, 2 moles of \( C_2H_6 \) will require:

\[ \text{Moles of } O_2 = 2 \text{ moles } C_2H_6 \times \frac{7}{2} \text{ moles } O_2 = 7 \text{ moles } O_2 \]

### Step 5: Convert moles of \( O_2 \) to liters at STP.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of \( O_2 \) required 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 \( C_2H_6 \) at STP is **156.8 liters**.

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