`a,n_{hh}=(8,96)/(22,4)=0,4(mol)`

PTHH:

`C_2H_4+Br_2->C_2H_4Br_2`

Vì chỉ có `C_2H_4` phản ứng

`->m_{C_2H_4}=m_{tăng}=3,36(g)`

`->n_{C_2H_4}=(3,36)/(28)=0,12(mol)`

Phần trăm số mol cũng là phần trăm thể tích

`\%C_2H_4=(0,12)/(0,4).100\%=30\%`

`->\%CH_4=(100-30)\%=70\%`

`b,`

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

`n_{C_2H_4}=30\%.0,1=0,03(mol)`

`->n_{CH_4}=0,07(mol)`

PTHH"

`C_2H_4+3O_2\overset{t^{o} }{\rightarrow}2CO_2+2H_2O`

`0,03->0,09(mol)`

`CH_4+2O_2\overset{t^{o} }{\rightarrow}CO_2+2H_2O`

`0,07->0,14(mol)`

`->n_{O_2}=0,09+0,14=0,23(mol)`

`->V_{O_2}=0,23.22,4=5,152(lít)`

`->V_{kk}=5,152.5=25,76(lít)`

Đáp án:

$a,\\ C_2H_4+Br_2\to C_2H_4Br_2\\ \%V_{CH_4}=70\%\\ \%V_{C_2H_4}=30\%\\ b,\ V_{kk}=25,76(l)$ 

Giải thích các bước giải:

$a,\\ C_2H_4+Br_2\to C_2H_4Br_2\\ m_{\text{bình brom tăng}}=m_{C_2H_4}=3,36(g)\\ n_{C_2H_4}=\dfrac{3,36}{28}=0,12(mol)\\ V_{C_2H_4}=0,12.22,4=2,688(l)\\ V_{CH_4}=8,96-2,688=6,272(l)\\ \%V_{CH_4}=\dfrac{6,272}{89,6}.100\%=70\%\\ \%V_{C_2H_4}=\dfrac{2,688}{8,96}.100\%=30\%\\ b,\\ \text{Trong 2,24 (l)}:\\ V_{CH_4}=2,24.70\%=1,568(l)\\ V_{C_2H_4}=2,24.30\%=0,672(l)\\ n_{CH_4}=\dfrac{1,568}{22,4}=0,07(mol)\\ n_{C_2H_4}=\dfrac{0,672}{22,4}=0,03(mol)\\ CH_4+2O_2\xrightarrow{t^0} CO_2+2H_2O\\ C_2H_4+3O_2\xrightarrow{t^0} 2CO_2+2H_2O\\ \xrightarrow{\text{Từ PTHH}}\ n_{O_2}=2.n_{CH_4}+3.n_{C_2H_4}=0,07.2+0,03.3=0,23(mol)\\ V_{O_2}=0,23.22,4=5,152(l)\\ V_{kk}=\dfrac{V_{O_2}}{20\%}=25,76(l)$