$PTPƯ:$
$Zn+2HCl\xrightarrow{} ZnCl_2+H_2↑$ $(1)$
$Al_2O_3+6HCl\xrightarrow{} 2AlCl_3+3H_2O$ $(2)$
$n_{H_2}=\dfrac{4,48}{22,4}=0,2mol.$
$Theo$ $pt1:$ $n_{Zn}=n_{H_2}=0,2mol.$
$⇒m_{Zn}=0,2.65=13g.$
$⇒m_{Al_2O_3}=23,2-13=10,2g.$
$⇒n_{Al_2O_3}=\dfrac{10,2}{102}=0,1mol.$
$Theo$ $pt1:$ $n_{ZnCl_2}=n_{H_2}=0,2mol.$
$Theo$ $pt2:$ $n_{AlCl_3}=2n_{Al_2O_3}=0,2mol.$
$⇒\%m_{ZnCl_2}=\dfrac{0,2.136}{0,2.136+0,2.133,5}.100\%=50,46\%$
$⇒\%m_{Al_2O_3}=\dfrac{0,2.133.5}{0,2.136+0,2.133,5}.100\%=49,54\%$
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