Bài 1 :

`n_{Br_2}=8/(160)=0,05(mol)`

PTHH:
`C_2H_4+Br_2\to C_2H_4Br_2`

`0,05\leftarrow 0,05(mol)`

`=>n_{C_2H_4}=0,05(mol)`

`->` Khí bay ra là methane

PTHH:

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

`Ba(OH)_2+CO_2\to BaCO_3+H_2O`

`n_{CH_4}=n_{CO_2}=n_{BaCO_3}=(29,55)/(197)=0,15(mol)`

`=>m=0,05.28+0,15.16=3,8(g)`

Bài 2 :

Gọi `n_{CH_4}=n_{C_2H_4}=a(mol)`

PTHH:

$\rm CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O \\ C_2H_4+3O_2 \xrightarrow{t^o} 2CO_2+2H_2O\\ Ca(OH)_2+CO_2\to CaCO_3+H_2O$

`=>n_{CO_2}=n_{CaCO_3}=3/(100)=0,03(mol)`

`=>n_{CH_4}+2n_{C_2H_4}=a+2a=0,03`

`=>a=0,01`

`=>V=a+a=0,02`

`=>V=0,02.24,79=0,4958(l)`

Bài 3 :

`a,`

`M_A=1,25.32=40(g//mol)`

`M_H=40.10\%=4`

`->H=4/1=4`

`M_C=40-4=36`

`->C=(36)/12=3`

`=>` CTPT là `C_3H_4`

CTCT :

$\rm \hspace{2.9cm}H\\\hspace{3cm}|\\H-C≡C-C-H\\\hspace{3cm}|\\\hspace{2.9cm}H$

`b,` 

`m=60.70\%=42(kg)`