$PTPƯ:$
$Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$ $(1)$
$2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$ $(2)$
$n_{Mg}=\dfrac{4,8}{24}=0,2mol$
$Theo$ $pt1:$ $n_{H_2}=n_{Mg}=0,2mol.$
$⇒V_{1}=0,2.22,4=4,48l.$ $(*)$
$n_{Al}=\dfrac{4,8}{27}=\dfrac{8}{45}mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{4}{15}mol.$
$⇒V_{2}=\dfrac{4}{15}.22,4=5,973l.$ $(**)$
$Từ$ $(*)$ và $(**)$ $⇒V_{1}<V_{2}$
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