$C_2H_4+Br_2\to C_2H_4Br_2\\ C_2H_2+2Br_2\to C_2H_2Br_4\\ n_{C_2H_4}:a(mol)\ ; n_{C_2H_2}:b(mol)\\ m_{tang}=m_{C_2H_4}+m_{C_2H_2}\\\to 28a+26b=8,2\ (1)\\ CH_4+2O_2\xrightarrow{t^o} CO_2+2H_2O\\ n_{CH_4}=n_{CO_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ n_{C_2H_4}+n_{C_2H_2}=\dfrac{10,08-0,15.22,4}{22,4}\\\to a+b=0,3\ (2)\\ (1)(2)\to a=0,2(mol)\ ; b=0,1(mol)\\ \%V_{C_2H_4}=\dfrac{0,2.22,4}{10,08}.100\%=44,44\%\\ \%V_{CH_4}=\dfrac{0,15.22,4}{10,08}.100\%=33,33\%\\ \%V_{C_2H_2}=100\%-33,33\%-44,44\%=22,23\%$

Đáp án:

Giải thích các bước giải:

` C_2H_4+Br_2->C_2H_4 Br_2`

`C_2H_2+2Br_2->C_2H_2Br_4`

$CH_4+2O_2\xrightarrow{t^o} CO_2+2H_2O$

`n_(CO_2)=n_(CH_4)=(3,36)/(22,4)=0,15(mol)`

`%V_(CH_4)=(0,15.22,4)/(10,08) .100%=33,33%`

Đặt `n_(C_2H_4)=x(mol)`

`n_(C_2H_2)=y(mol)`

`n_(C_2H_4)+n_(C_2H_2)=x+y=(10,08-3,36)/(22,4)=0,3(mol)`

`m_(tăng)=m_(C_2H_4)+m_(C_2H_2)=28x+26y=8,2(g)`

giải hệ PT

`=>x=0,2;y=0,1`

`%V_(C_2H_4)=(0,2.22,4)/(10,08)=44,44%`

`%V_(C_2H_2)=22,23%`