$a,PTPƯ:Zn+2HCl\xrightarrow{} ZnCl_2+H_2↑$
$b,n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2mol.$
$Theo$ $pt:$ $n_{Zn}=\dfrac{1}{2}n_{HCl}=0,1mol.$
$⇒a=m_{Zn}=0,1.65=6,5g.$
$c,Theo$ $pt:$ $n_{ZnCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1mol.$
$⇒C\%_{ZnCl_2}=\dfrac{0,1.136}{6,5+100-(0,1.2)}.100\%=12,79\%$
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