`a)`

`n_{C_2H_4Br_2}={18,8}/{188}=0,1(mol)`

`C_2H_4+Br_2->C_2H_4Br_2`

`0,1←0,1←0,1(mol)`

`m_{Br_2}=0,1.160=16(g)`

`V_{C_2H_4}=0,1.22,4=2,24(L)`

`b)`

`C_2H_4+3O_2` $\xrightarrow{t^o}$ `2CO_2+2H_2O`

`0,1->0,3(mol)`

`V_{kk}={0,3.22,4}/{20\%}=33,6(L)`

`\color{red}{#aether}`

`a.`

PTHH : `C_2H_4+Br_2->C_2H_4Br_2`

`n_{C_2H_4Br_2}=(18,8)/(188)=0,1` `mol`

Theo phương trình : `n_{Br_2}=n_{C_2H_4}=n_{C_2H_4Br_2}=0,1` `mol`

`=>m_{Br_2}=0,1.160=16` `g`

`V_{C_2H_4}=0,1.22,4=2,24` `l`

`b.`

PTHH : `C_2H_4+3O_2\overset(t^o)(->)2CO_2+2H_2O`

Theo phương trình : `n_{O_2}=3n_{C_2H_4}=0,3` `mol`

`=>V_{O_2}=0,3.22,4=6,72` `l`

`%V_{O_2}=20%V_{kk}`

`=>V_{kk}=6,72:20%=33,6` `l`