$2H_2+O_2\buildrel{{t^o}}\over\to 2H_2O$
Theo lí thuyết, $O_2$ hết.
Thực tế $H=75\%$ nên $n_{O_2\text{pứ}}=0,08.75\%=0,06(mol)$
$\Rightarrow n_{H_2O}=n_{H_2\text{pứ}}=0,06.2=0,12(mol)$
$m_{H_2O}=0,12.18=2,16g$
$m_{H_2\text{dư}}=2(0,2-0,12)=0,16g$
$m_{O_2\text{dư}}=32(0,08-0,06)=0,64g$