\(CH_4+2O_2\underrightarrow{^{to}}CO_2+2H_2O\\ n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\\ n_{CH_4\left(LT\right)}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ n_{CH_4\left(TT\right)}=0,1:\left(100\%-10\%\right)=\dfrac{1}{9}\left(mol\right)\\ V_{CH_4\left(TT\right)}=\dfrac{1}{9}.22,4\approx2,489\left(l\right)\)