`EtNH_2 + MeMgI overset("Heated at high temp.") underset("in the presence of pyri di ne ")rarr Gas (A)` The volume of gas `(A)` obtained at `S.T.P` when `0.45 gm` of `EtNH_2` reacts with `MeMgI` is.

Options

A. `224 ml`
B. `22.4 ml`
C. `448 ml`
D. 44.8 ml`

Explanation

To solve the problem, we need to determine the volume of gas (methane, CH₄) produced when 0.45 grams of ethylamine (EtNH₂) reacts with methyl magnesium iodide (MeMgI) under the specified conditions.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between ethylamine (EtNH₂) and methyl magnesium iodide (MeMgI) at high temperature in the presence of pyridine produces methane (CH₄) and a byproduct.

2. **Calculate the Molar Mass of Ethylamine**:
- The molecular formula of ethylamine is C₂H₅NH₂.
- Molar mass calculation:
- Carbon (C): 2 × 12 g/mol = 24 g/mol
- Hydrogen (H): 7 × 1 g/mol = 7 g/mol
- Nitrogen (N): 1 × 14 g/mol = 14 g/mol
- Total = 24 + 7 + 14 = 45 g/mol

3. **Determine Moles of Ethylamine**:
- Using the formula:
\[ \text{Moles of EtNH}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.45 \text{ g}}{45 \text{ g/mol}} = 0.01 \text{ moles} \]

4. **Stoichiometry of the Reaction**:
- The reaction produces 1 mole of methane (CH₄) for every mole of ethylamine (EtNH₂) consumed.
- Therefore, 0.01 moles of ethylamine will produce 0.01 moles of methane.

5. **Calculate the Volume of Methane at STP**:
- At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters.
- Volume of methane produced:
\[ \text{Volume of CH}_4 = \text{moles of CH}_4 \times 22.4 \text{ L/mol} = 0.01 \text{ moles} \times 22.4 \text{ L/mol} = 0.224 \text{ L} \]
- Convert to milliliters:
\[ 0.224 \text{ L} = 224 \text{ mL} \]

6. **Final Answer**:
The volume of gas (A) obtained at STP when 0.45 grams of ethylamine reacts with methyl magnesium iodide is **224 mL** of methane (CH₄).