Correct option is (2) 0.25
\( \mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+2}+\frac{3 \mathrm{n}+1}{2} \mathrm{O}_2 \longrightarrow \mathrm{nCO}_2+(\mathrm{n}+1) \mathrm{H}_2 \mathrm{O} \)
\(\underset{\underset {0.25\,mole}{4gm}}{CH_4}+2O_2 \rightarrow \underset {\underset {0.25\,mole}{11gm}}{CO_2} + 2H_2O\)
\(0.25 \) mole \(CH_4 \) gives 0.25 mole (or 11 gm) \(CO_2\)