$a,PTPƯ:2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$
$n_{Al}=\dfrac{2,7}{27}=0,1mol.$
$Theo$ $pt:$ $n_{HCl}=3n_{Al}=0,3mol$
$⇒m_{HCl}=0,3.36,5=10,95g.$
$⇒m_{ddHCl}=\dfrac{10,95}{20\%}=54,75g.$
$b,Theo$ $pt:$ $n_{AlCl_3}=n_{Al}=0,1mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=0,15mol.$
$⇒C\%_{AlCl_3}=\dfrac{0,1.133,5}{2,7+54,75-(0,15.2)}.100\%=23.36\%$
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