$a,PTPƯ:2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$
$b,n_{Al}=\dfrac{10,8}{ 27}=0,4mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=0,6mol.$
$⇒V_{H_2}=0,6.22,4=13,44l.$
$c,PTPƯ:CuO+H_2\xrightarrow{t^o} Cu+H_2O$
$n_{CuO}=\dfrac{10}{80}=0,125mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,125}{1}<\dfrac{0,6}{1}$
$⇒H_2$ $dư.$
$Theo$ $pt:$ $n_{Cu}=n_{CuO}=0,125mol.$
$⇒m_{Cu}=0,125.64=8g.$
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