Equal masses of `H_(2)`, `O_(2)` and methane have been taken in a container of volume `V` at temperature `27^(@)C` in identical conditions. The ratio of the volume of gases `H_(2) : O_(2)` : methane would be

Options

A. `8 : 16 : 1`
B. `16 : 8 :1`
C. `16 : 1 :2`
D. `8 : 1 : 2`

Explanation

To solve the problem of finding the ratio of the volumes of gases \( H_2 \), \( O_2 \), and methane (\( CH_4 \)) when equal masses of each gas are taken in a container at constant temperature and pressure, we can follow these steps:

### Step 1: Understand the Conditions
We are given equal masses of \( H_2 \), \( O_2 \), and \( CH_4 \) at a constant temperature of \( 27^\circ C \) and in identical conditions. According to Avogadro's law, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of that gas.

### Step 2: Define the Masses
Let the equal mass of each gas be \( W \).

### Step 3: Calculate the Number of Moles
The number of moles \( n \) of each gas can be calculated using the formula:
\[ n = \frac{W}{M} \]
where \( M \) is the molar mass of the gas.

- For \( H_2 \):
\[ n_{H_2} = \frac{W}{2} \quad (\text{Molar mass of } H_2 = 2 \text{ g/mol}) \]

- For \( O_2 \):
\[ n_{O_2} = \frac{W}{32} \quad (\text{Molar mass of } O_2 = 32 \text{ g/mol}) \]

- For \( CH_4 \):
\[ n_{CH_4} = \frac{W}{16} \quad (\text{Molar mass of } CH_4 = 16 \text{ g/mol}) \]

### Step 4: Set Up the Volume Ratio
According to Avogadro's law, the volume ratio is the same as the mole ratio:
\[ V_{H_2} : V_{O_2} : V_{CH_4} = n_{H_2} : n_{O_2} : n_{CH_4} \]

Substituting the values we calculated:
\[ V_{H_2} : V_{O_2} : V_{CH_4} = \frac{W}{2} : \frac{W}{32} : \frac{W}{16} \]

### Step 5: Simplify the Ratios
To simplify, we can eliminate \( W \) from all terms:
\[ \frac{1}{2} : \frac{1}{32} : \frac{1}{16} \]

Now, we can find a common denominator to simplify further. The least common multiple of 2, 32, and 16 is 32. Therefore, we convert each term:
\[ \frac{1}{2} = \frac{16}{32}, \quad \frac{1}{32} = \frac{1}{32}, \quad \frac{1}{16} = \frac{2}{32} \]

So, the ratio becomes:
\[ 16 : 1 : 2 \]

### Step 6: Final Ratio
Thus, the final ratio of the volumes of the gases \( H_2 : O_2 : CH_4 \) is:
\[ \text{Ratio} = 16 : 1 : 2 \]

### Conclusion
The correct answer is \( 16 : 1 : 2 \).

---