$a,PTPƯ:Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$
$n_{Mg}=\dfrac{2,4}{24}=0,1mol.$
$Theo$ $pt:$ $n_{HCl}=2n_{Mg}=0,2mol.$
$⇒m_{ddHCl}=\dfrac{0,2.36,5}{14,6\%}=50g.$
$b,Theo$ $pt:$ $n_{MgCl_2}=n_{H_2}=n_{Mg}=0,1mol.$
$⇒C\%_{MgCl_2}=\dfrac{0,1.95}{2,4+50-(0,1.2)}.100\%=18,2\%$
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