$a,PTPƯ:Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$
$b,n_{Mg}=\dfrac{4,8}{24}=0,2mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Mg}=0,2mol.$
$⇒V_{H_2}=0,2.22,4=4,48l.$
$c,Theo$ $pt:$ $n_{HCl}=2n_{Mg}=0,4mol.$
$⇒m_{HCl}=\dfrac{0,4.36,5}{25\%}=58,4g.$
$d,Theo$ $pt:$ $n_{MgCl_2}=n_{Mg}=0,2mol.$
$⇒CM_{MgCl_2}=\dfrac{0,2.95}{4,8+58,4-(0,2.2)}.100\%=30,25\%$
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