d
Let the mass of \(\mathrm{H}_{2}\) gas be \(\mathrm{x} \mathrm{ g}\) and mass of \(\mathrm{O}_{2}\) gas \(4 \mathrm{xg}\) Molar

\(\quad \quad \mathrm{H}_{2}: \mathrm{O}_{2}\)

mass \(2: 32\)

i.e. \(1: 16\)

therefore, Molar ratio \(=\frac{n_{\mathrm{H}_{12}}}{n_{0_{2}}}=\frac{x / 2}{4 x / 32}=\frac{(x) \times 32}{2 \times 4 x}=\frac{4}{1}=4: 1\)