$a/$
$C_2H_4 + 2O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O$
$C_3H_6 + \dfrac{9}{2}O_2 \xrightarrow{t^o} 3CO_2 + 3H_2O$
$b/$
$n_{O_2} = \dfrac{23,52}{22,4} = 1,05(mol)$
Gọi $n_{C_2H_4} = a(mol) ; n_{C_3H_6} = b(mol)$
$\to 28a + 42b = 9,8(1)$
Theo ` PTHH : `
$n_{O_2} = 2a + 4,5b = 1,05(2)$
Từ `(1)(2)` suy ra $a = 0 \to$ Bạn xem lại đề