`n_{CH_4}={5,6}/{22,4}=0,25(mol)`

`n_{O_2}={48}/{32}=1,5(mol)`

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

`0,25->0,5->0,25->0,5(mol)`

`0,25<{1,5}/2->O_2` dư

`m_{CO_2}=0,25.44=22(g)`

`m_{H_2O}=0,5.18=9(g)`

`m_{O_2\ du}=(1,5-0,5).32=32(g)`

$PTHH : CH_4 +2O_2 \xrightarrow{to} CO_2 +2H_2O$

$n_{CH_4}=\dfrac{5,6}{22,4}=0,25(mol)$

$n_{O_2}=\dfrac{48}{32}=1,5(mol)$

tỉ lệ $\dfrac{0,25}{1} <\dfrac{1,5}{2}$ ⇒ $CH_4$ hết 

$n_{CO_2}= n_{CH_4}=0,25(mol)$

→ $m_{CO_2}=0,25 ×44=11(g)$

$n_{H_2O}=2n_{CH_4}=0,25×2=0,5(mol)$

→$m_{H_2O}=0,5×18 =9(g)$

$n_{O_2 pư}=0,5 (mol)$

→$m_{O_2 pư}=0,5×32 =16(g)$

⇒$m_{O_2 dư}=48-16=32(g)$