$a,PTPƯ:2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$
$b,n_{Al}=\dfrac{8,1}{27}=0,3mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=0,45mol$
$⇒V_{H_2}=0,45.22,4=10,08l.$
$c,PTPƯ:CuO+H_2\xrightarrow{t^o} Cu+H_2O$
$n_{CuO}=\dfrac{12}{80}=0,15mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,15}{1}<\dfrac{0,45}{1}$
$⇒H_2$ $dư.$
$⇒n_{H_2}(dư)=0,45-\dfrac{0,15.1}{1}=0,3mol.$
$⇒m_{H_2}(dư)=0,3.2=0,6g.$
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