A container has an equal mass of H₂, O₂ and CH₄ at 27℃, the ratio of their volume is

The correct option is 16:1:2.

To understand why this is the correct ratio, we can apply Avogadro's Law and the Ideal Gas Law. Under identical conditions of temperature (27℃) and pressure, the volume ($V$) of an ideal gas is directly proportional to the number of moles ($n$) of the gas.
This relation is given by:
$V\propto n$

Therefore, the ratio of the volumes of the gases is equal to the ratio of their number of moles:
$V_{H_2}:V_{O_2}:V_{C{H}_4}=n_{H_2}:n_{O_2}:n_{C{H}_4}$

Let the mass of each gas in the container be $w$ grams (since the container has an equal mass of each gas).
The molar masses (molecular weights) of the respective gases are:
• Hydrogen ($H_2$) = 2 g/mol
• Oxygen ($O_2$) = 32 g/mol
• Methane ($C{H}_4$) = 12 + 4(1) = 16 g/mol

Using the formula for the number of moles, which is mass divided by molar mass ($n=\frac{w}{M}$), we get the number of moles for each gas:
• Moles of Hydrogen, $n_{H_2}=\frac{w}{2}$
• Moles of Oxygen, $n_{O_2}=\frac{w}{32}$
• Moles of Methane, $n_{C{H}_4}=\frac{w}{16}$

Now, we substitute these expressions into the volume ratio:
$V_{H_2}:V_{O_2}:V_{C{H}_4}=\frac{w}{2}:\frac{w}{32}:\frac{w}{16}$

We can divide each term by the common mass variable $w$ to simplify the ratio:
$V_{H_2}:V_{O_2}:V_{C{H}_4}=\frac{1}{2}:\frac{1}{32}:\frac{1}{16}$

To convert this fractional ratio into a whole number ratio, we multiply each term by the least common multiple (LCM) of the denominators (2, 32, and 16), which is 32:
• First term: $\frac{1}{2}\times 32=16$
• Second term: $\frac{1}{32}\times 32=1$
• Third term: $\frac{1}{16}\times 32=2$

This gives us the final simplified volume ratio:
$V_{H_2}:V_{O_2}:V_{C{H}_4}=16:1:2$