$PTHH : C_2H_4 +3O_2 \xrightarrow{to} 2CO_2 +2H_2O$ 

$n_{O_2}= \dfrac{3,36}{22,4}=0,15(mol)$ 

Theo pt 

$n_{C_2H_4}=\dfrac{1}{3}n_{O_2}=0,05(mol)$ 

$\to V_{C_2H_4}=0,05 ×22,4 =1,12(lít)$ 

`\to %V_{C_2H_4} ={1,12}/{4,48}×100%=25%`

`=> %V_{CO_2}=100%-25%=75%`

`a.`

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

`b.`

`n_{O_2}=(3,36)/(22,4)=0,15` `mol`

Theo phương trình : `n_{C_2H_4}=(1)/(3)n_{O_2}=0,05` `mol`

`=>V_{C_2H_4}=0,05.22,4=1,12` `l`

`=>%C_2H_4=(1,12)/(4,48).100%=25%`

`=>%CO_2=100%-25%=75%`