A compound with formula `C_(6)H_(6)O_(2)` gave methane on treatment with excess of MeMgBr, 5.5 g of this compound gave 2.24 litres of methane at STP. How many active hydrogens are present in the compound.

Answer

To solve the problem, we need to determine how many active hydrogens are present in the compound with the formula \( C_6H_6O_2 \). The compound produces methane when treated with excess methylmagnesium bromide (MeMgBr).

### Step-by-Step Solution:

1. **Calculate the Molar Mass of the Compound**:
The molar mass of \( C_6H_6O_2 \) can be calculated as follows:
- Carbon (C): \( 6 \times 12 = 72 \, \text{g/mol} \)
- Hydrogen (H): \( 6 \times 1 = 6 \, \text{g/mol} \)
- Oxygen (O): \( 2 \times 16 = 32 \, \text{g/mol} \)
- Total molar mass = \( 72 + 6 + 32 = 110 \, \text{g/mol} \)

2. **Determine the Number of Moles of the Compound**:
Given that 5.5 g of the compound is used, we can find the number of moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.5 \, \text{g}}{110 \, \text{g/mol}} = 0.05 \, \text{mol} \]

3. **Calculate the Volume of Methane Produced**:
The problem states that 5.5 g of the compound produced 2.24 liters of methane at STP. At STP, 1 mole of gas occupies 22.4 liters. Therefore, we can calculate the number of moles of methane produced:
\[ \text{Number of moles of methane} = \frac{2.24 \, \text{L}}{22.4 \, \text{L/mol}} = 0.1 \, \text{mol} \]

4. **Relate Moles of Compound to Moles of Methane**:
From the reaction, we know that 1 mole of \( C_6H_6O_2 \) gives 2 moles of methane. Therefore, if we produced 0.1 moles of methane, we can find the moles of \( C_6H_6O_2 \) that reacted:
\[ \text{Moles of } C_6H_6O_2 = \frac{0.1 \, \text{mol of methane}}{2} = 0.05 \, \text{mol} \]

5. **Determine the Number of Active Hydrogens**:
Since we have established that 0.05 moles of \( C_6H_6O_2 \) produced 0.1 moles of methane, it indicates that there are 2 active hydrogens in each molecule of the compound. Thus, the total number of active hydrogens in the compound is:
\[ \text{Active hydrogens} = 2 \times 0.05 \, \text{mol} = 0.1 \, \text{mol} \]

### Conclusion:
The compound \( C_6H_6O_2 \) contains **2 active hydrogens**.

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