$a,PTPƯ:2H_2+O_2\xrightarrow{t^o} 2H_2O$
$n_{H_2}=\dfrac{3,36}{22,4}=0,15mol.$
$n_{O_2}=\dfrac{11,2}{22,4}=0,5mol.$
$⇒O_2$ $dư.$
$⇒n_{O_2}(dư)=0,5-\dfrac{0,15.1}{2}=0,425mol.$
$⇒m_{O_2}(dư)=0,425.32=13,6g.$
$b,Theo$ $pt:$ $n_{H_2O}=n_{H_2}=0,15mol.$
$⇒m_{H_2O}=0,15.18=2,7g.$
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