$11,3(g)\ X: \begin{cases} n_{CH_4}:a(mol)\\n_{C_2H_4}:b(mol)\\n_{C_2H_2}:c(mol) \end{cases}\\ m_X=16a+28b+26c=11,3\ (1)\\ Q_{toa}=783a+1423b+1300c=566,13\ (2)\\ C_2H_4+Br_2\to C_2H_4Br_2\\ C_2H_2+2Br_2\to C_2H_2Br_4\\ n_{Br_2}=\dfrac{80}{160}=0,5(mol)\\ \sum n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}\\\to b+2c=0,5\ (3)\\ (1)(2)(3)\to \begin{cases} a=0,15(mol)\\b=0,16(mol)\\c=0,17(mol) \end{cases}\\ \%V_{CH_4}=\dfrac{0,15}{0,15+0,16+0,17}.100\%=31,25\%$

`X:CH_4:x(mol);C_2H_4:y(mol);C_2H_2:z(mol)`

`->16x+28y+26z=11,3(1)`

`Q_{toa}=783x+1423y+1300z=566,13(2)`

`C_2H_4+Br_2->C_2H_4Br_2`

`C_2H_2+2Br_2->C_2H_2Br_4`

`->n_{Br_2}=y+2z={80}/{160}=0,5(3)`

`(1)(2)(3)->x=0,15;y=0,16;z=0,17`

`->\%V_{CH_4}={0,15}/{0,15+0,16+0,17}.100\%=31,25\%`