`n_{\text{hh khí}}=\frac{4,48}{22,4}=0,2(mol)`
`n_{CH_4}=\frac{3,36}{22,4}=0,15(mol)`
`=> n_{C_2H_4}=0,2-0,15=0,05(mol)`
`a)` `CH_2=CH_2+Br_2\to CH_2Br-CH_2Br`
`%m_{C_2H_4}=\frac{28.0,05}{28.0,05+16.0,15}.100%\approx36,84%`
`%m_{CH_4}=\frac{16.0,15}{3,8}.100%\approx 63,16%`
`b)` `CH_4+2O_2\overset{t^o}{\to}CO_2+2H_2O`
`C_2H_4+3O_2\overset{t^o}{\to}2CO_2+2H_2O`
`n_{O_2}=2n_{CH_2}+3n_{C_2H_4}=0,45(mol)`
`=> V_{(kk)}=5.V_{O_2}`
`=> V_{(kk)}=5.22,4.0,45=50,4(l)`