$a,PTPƯ:Zn+2HCl\xrightarrow{} ZnCl_2+H_2↑$
$b,n_{H_2}=\dfrac{4,48}{22,4}=0,2mol.$
$Theo$ $pt:$ $n_{Zn}=n_{H_2}=0,2mol.$
$⇒m=m_{Zn}=0,2.65=13g.$
$c,Theo$ $pt:$ $n_{HCl}=2n_{H_2}=0,4mol.$
$⇒C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%$
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