At STP , for complete combustion of 3 g `C_(2)H_(6)` the required volume of `O_(2)` will be -

Options

A. 78. 4 L
B. 7. 84 L
C. 2. 78 L
D. 6. 23 L

Answer

To solve the question of how much volume of \( O_2 \) is required for the complete combustion of 3 g of \( C_2H_6 \) (ethane) at STP, we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of \( C_2H_6 \).
The balanced equation for the combustion of ethane is:
\[ C_2H_6 + \frac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O \]

### Step 2: Determine the molar mass of \( C_2H_6 \).
The molar mass of \( C_2H_6 \) can be 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.

Calculating the molar mass:
\[ \text{Molar mass of } C_2H_6 = (2 \times 12) + (6 \times 1) = 24 + 6 = 30 \text{ g/mol} \]

### Step 3: Calculate the number of moles of \( C_2H_6 \) in 3 g.
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{Number of moles of } C_2H_6 = \frac{3 \text{ g}}{30 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 4: Use the stoichiometry of the balanced equation to find moles of \( O_2 \) required.
From the balanced equation, 1 mole of \( C_2H_6 \) requires \( \frac{7}{2} \) moles of \( O_2 \). Therefore, for 0.1 moles of \( C_2H_6 \):
\[ \text{Moles of } O_2 = 0.1 \times \frac{7}{2} = 0.35 \text{ moles} \]

### Step 5: Calculate the volume of \( O_2 \) 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:
\[ \text{Volume of } O_2 = 0.35 \text{ moles} \times 22.4 \text{ L/mol} = 7.84 \text{ L} \]

### Final Answer:
The required volume of \( O_2 \) for the complete combustion of 3 g of \( C_2H_6 \) at STP is **7.84 liters**.

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