$a,PTPƯ:Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$
$b,n_{Mg}=\dfrac{4,8}{24}=0,2mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Mg}=0,2mol.$
$⇒V_{H_2}=0,2.22,4=4,48l.$
$c,PTPƯ:2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$
$Theo$ $pt:$ $n_{Na}=2n_{H_2}=0,4mol.$
$⇒m_{Na}=0,4.23=9,2g.$
$d,PTPƯ:Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$
$Theo$ $pt:$ $n_{HCl}=2n_{Mg}=0,4mol.$
$⇒m_{HCl}=0,4.36,5=14,6g.$
$⇒m_{ddHCl}=\dfrac{14,6}{7,3\%}=200g.$
Vì axit lấy dư 10% so với lí thuyết nên:
$⇒m_{ddHCl}=200.(100\%+10\%)=220g.$
chúc bạn học tốt!