$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$n_{Fe}=\dfrac{6}{56}=\dfrac{3}{28}mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Fe}=\dfrac{3}{28}mol.$
$⇒V_{H_2}=\dfrac{3}{28}.22,4=2,4l.$
$b,Theo$ $pt:$ $n_{HCl}=2n_{Fe}=\dfrac{3}{14}mol.$
$⇒m_{HCl}=\dfrac{3}{14}.36,5=7,82g.$
$c,PTPƯ:MgO+H_2\xrightarrow{t^o} Mg+H_2O$
$n_{MgO}=\dfrac{16}{40}=0,4mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,4}{1}>\dfrac{\frac{3}{28}}{1}$
$⇒MgO$ $dư.$
$Theo$ $pt:$ $n_{Mg}=n_{H_2}=\dfrac{3}{28}mol.$
$⇒m_{Mg}=\dfrac{3}{28}.24=2,57g.$
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