$a,PTPƯ:Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$
$n_{Mg}=\dfrac{4,8}{24}=0,2mol.$
Đổi 100 ml = 0,1 lít.
$⇒n_{HCl}=0,1.2=0,2mol.$
$⇒Mg$ $dư.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1mol.$
$⇒V_{H_2}=0,1.22,4=2,24l.$
$b,Mg$ $dư.$
$⇒n_{Mg}(dư)=0,2-\dfrac{0,2.1}{2}=0,1mol$
$⇒m_{Mg}(dư)=0,1.24=2,4g.$
$c,Theo$ $pt:$ $n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,1mol.$
$⇒CM_{MgCl_2}=\dfrac{0,1}{0,1}=1M.$
chúc bạn học tốt!