The volume of methane at N.T.P formed from 8.2 g of sodium acetate by fusion with soda lime is

Options

A. 10 L
B. 11.2 L
C. 5.6 L
D. 2.24 L

Explanation

To solve the problem of finding the volume of methane (CH₄) formed from 8.2 g of sodium acetate (CH₃COONa) by fusion with soda lime, we can follow these steps:

### Step 1: Write the Reaction
The reaction of sodium acetate with soda lime (a mixture of sodium hydroxide and calcium oxide) can be represented as follows:
\[ \text{CH}_3\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO}} \text{CH}_4 + \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \]
This indicates that one mole of sodium acetate produces one mole of methane.

### Step 2: Calculate the Molar Mass of Sodium Acetate
The molar mass of sodium acetate (CH₃COONa) can be calculated as follows:
- Carbon (C): 12 g/mol × 2 = 24 g/mol
- Hydrogen (H): 1 g/mol × 3 = 3 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol
- Sodium (Na): 23 g/mol × 1 = 23 g/mol

Adding these together:
\[ \text{Molar mass of CH}_3\text{COONa} = 24 + 3 + 32 + 23 = 82 \text{ g/mol} \]

### Step 3: Calculate the Number of Moles of Sodium Acetate
Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{Given weight}}{\text{Molar mass}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{8.2 \text{ g}}{82 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 4: Calculate the Volume of Methane at NTP
At Normal Temperature and Pressure (NTP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of methane produced from 0.1 moles can be calculated as:
\[ \text{Volume} = \text{Number of moles} \times 22.4 \text{ L/mol} \]
Substituting the values:
\[ \text{Volume} = 0.1 \text{ moles} \times 22.4 \text{ L/mol} = 2.24 \text{ liters} \]

### Final Answer
The volume of methane formed from 8.2 g of sodium acetate at NTP is **2.24 liters**.

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