$a,PTPƯ:Zn+2HCl\xrightarrow{} ZnCl_2+H_2↑$
b, Đổi 200 ml = 0,2 lít.
$⇒n_{HCl}=0,2.2=0,4mol.$
$Theo$ $pt:$ $n_{Zn}=\dfrac{1}{2}n_{HCl}=0,2mol.$
$⇒m_{Zn}=0,2.65=13g.$
$c,Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{HCl}=0,2mol.$
$⇒V_{H_2}=0,2.22,4=4,48l.$
$d,Theo$ $pt:$ $n_{ZnCl_2}=\dfrac{1}{2}n_{HCl}=0,2mol.$
$⇒CM_{ZnCl_2}=\dfrac{0,2}{0,2}=1M.$
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