1/
$Fe_2O_3+6HCl\xrightarrow{} 2FeCl_3+3H_2O$
2/
$a,PTPƯ:Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$
b, Đổi 150 ml = 0,15 lít.
$n_{HCl}=0,15.1=0,15mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{HCl}=0,075mol.$
$⇒V_{H_2}=0,075.22,4=1,68l.$
$c,Theo$ $pt:$ $n_{Mg}=\dfrac{1}{2}n_{HCl}=0,075mol.$
$⇒m_{Mg}=0,075.24=1,8g.$
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