khí thoát ra là: `CH_4`

`a.%V_{CH_4}={2,24}/{6,72}.100%~~33,33%`

`%V_{C_2H_4}=100-33,33=66,67%`

`b.n_{C_2H_4}={6,72-2,24}/{22,4}=0,2(mol)`

`n_{CH_4}={2,24}/{22,4}=0,1(mol)`

`PTHH:CH_4+3O_2 \overset(t^o)(->)CO_2+2H_2O`

`(mol) ``0,1->`     `0,3`

          `2C_2H_2+5O_2 \overset(t^o)(->)4CO_2+2H_2O`

`(mol)`  `0,2->`       `0,5`

`n_{O_2}=0,3+0,5=0,8(mol)`

`=>V_{O_2}=0,8.22,4=17,92(l)`

$a)\\ C_2H_2+2Br_2\xrightarrow\ C_2H_2Br_4\\ \%V_{CH_4}=\dfrac{2,24}{6,72}.100\%=33,33\%\\ \%V_{C_2H_2}=100\%-33,33\%=66,67\%\\ b)\\ CH_4+2O_2\xrightarrow{t^o} CO_2+2H_2O\\ 2C_2H_2+5O_2\xrightarrow{t^o} 4CO_2+2H_2O\\ n_{C_2H_2}=\dfrac{6,72-2,24}{22,4}=0,2(mol)\\ n_{CH_4}=\dfrac{2,24}{22,4}=0,1(mol)\\ \sum n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=0,7(mol)\\ V_{O_2}=0,7.22,4=15,68(l)$