a)
$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
$C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O$
b)
Gọi $V_{CH_4} = a ; V_{C_2H_4} = b$
Ta có :
$a + b = 5,6$
$V_{CO_2} = a + 2b = 8,96$
Suy ra a = 2,24 ; b = 3,36
Vậy :
$\%V_{CH_4} = \dfrac{2,24}{5,6}.100\% = 40\%$
$\%V_{C_2H_4} = 100\% -40\% = 60\%$
c)
$V_{O_2} = 2a + 3b = 2,24.2 + 3,36.3 = 14,56(lít)$