How many litres of methane would be produced when 0.6 gm of `Ch_(3)`MgBr is treated with excess of `C_(4)H_(9)NH_(2)`?

Options

A. 0.8 litre
B. 0.08 litre
C. 0.112 litre
D. 1.12 litre

Explanation

To solve the problem of how many liters of methane (CH₄) would be produced when 0.6 grams of CH₃MgBr is treated with an excess of C₄H₉NH₂, we can follow these steps:

### Step 1: Determine the molar mass of CH₃MgBr
- The molecular formula of CH₃MgBr consists of:
- Carbon (C): 1 atom × 12 g/mol = 12 g/mol
- Hydrogen (H): 3 atoms × 1 g/mol = 3 g/mol
- Magnesium (Mg): 1 atom × 24 g/mol = 24 g/mol
- Bromine (Br): 1 atom × 79 g/mol = 79 g/mol

**Calculation:**
\[ \text{Molar mass of CH}_3\text{MgBr} = 12 + 3 + 24 + 79 = 118 \text{ g/mol} \]

### Step 2: Calculate the number of moles of CH₃MgBr
- Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
- For 0.6 grams of CH₃MgBr:
\[ \text{Number of moles of CH}_3\text{MgBr} = \frac{0.6 \text{ g}}{118 \text{ g/mol}} \approx 0.00508 \text{ moles} \]

### Step 3: Determine the amount of methane produced
- The reaction of CH₃MgBr with C₄H₉NH₂ produces methane (CH₄) in a 1:1 molar ratio.
- Therefore, the number of moles of CH₄ produced is the same as the number of moles of CH₃MgBr:
\[ \text{Number of moles of CH}_4 = 0.00508 \text{ moles} \]

### Step 4: Calculate the volume of methane produced at STP
- At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters.
- Thus, the volume of CH₄ produced is:
\[ \text{Volume of CH}_4 = \text{Number of moles} \times 22.4 \text{ L/mol} \]
\[ \text{Volume of CH}_4 = 0.00508 \text{ moles} \times 22.4 \text{ L/mol} \approx 0.113 \text{ L} \]

### Final Answer
Approximately **0.113 liters** of methane (CH₄) would be produced.

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