$a,PTPƯ:2Al+3Cl_2\xrightarrow{t^o} 2AlCl_3$
$b,n_{Al}=\dfrac{5,4}{27}=0,2mol.$
$n_{Cl_2}=\dfrac{8,96}{22,4}=0,4mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,2}{2}<\dfrac{0,4}{3}$
$⇒Cl_2$ $dư.$
$⇒n_{Cl_2}(dư)=0,4-\dfrac{0,2.3}{2}=0,1mol.$
$⇒m_{Cl_2}(dư)=0,1.71=7,1g.$
$c,Theo$ $pt:$ $n_{AlCl_3}=n_{Al}=0,2mol.$
$⇒m_{AlCl_3}=0,2.133,5=26,7g.$
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