To solve this problem, we need to find the ratio of the volumes of $H_2, O_2,$ and methane when equal masses are taken.$\\$Given:$\\$• Equal masses of $H_2, O_2,$ and methane are taken.$\\$• Temperature $= 27^{\circ}C = 300 \,$K$\\$• Volume of the container $= V \\$Let's denote:$\\$• $M_{H_2} = 2 \,$g/mol (molar mass of $H_2) \\$• $M_{O_2} = 32 \,$g/mol (molar mass of $O_2) \\$• $M_{methane} = 16 \,$g/mol (molar mass of methane)$\\$Since the masses are equal, let the mass of each gas be $m. \\$Number of moles of each gas is given by $n = \frac{m}{M}. \\$Calculate the number of moles for each gas:$\\ n_{H_2} = \frac{m}{2} \\ n_{O_2} = \frac{m}{32} \\ n_{methane} = \frac{m}{16} \\$According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain equal numbers of moles.$\\$Therefore, the volume ratio is the same as the mole ratio:$\\$Volume ratio $= n_{H_2} : n_{O_2} : n_{methane} \\$Substitute the values:$\\ \frac{m}{2} : \frac{m}{32} : \frac{m}{16} \\$Simplify the ratio:$\\ \frac{1}{2} : \frac{1}{32} : \frac{1}{16} \\$To eliminate fractions, multiply each term by the least common multiple of the denominators (32):$\\ 16 : 1 : 2 \\$Thus, the ratio of the volumes of $H_2, O_2,$ and methane is $16 : 1 : 2. \\$This corresponds to Option 3.