$a,PTPƯ:2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$

$n_{Al}=\dfrac{12,15}{27}=0,45mol.$

$n_{HCl}=\dfrac{109,5.20\%}{36,5}=0,6mol.$

$\text{Lập tỉ lệ:}$ $\dfrac{0,45}{2}>\dfrac{0,6}{6}$

$⇒Al$ $dư.$

$⇒n_{Al}(dư)=0,45-\dfrac{0,6.2}{6}=0,25mol.$

$⇒m_{Al}(dư)=0,25.27=6,75g.$

$b,Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3mol.$

$Theo$ $pt:$ $n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,2mol.$

$⇒C\%_{AlCl_3}=\dfrac{0,2.133,5}{(12,15-6,75)+109,5-(0,3.2)}.100\%=23,36\%$

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$n_{Al}=12,15/27=0,45mol$

$m_{HCl}=109,5.20\%=21,9g$

$⇒n_{HCl}=21,9/36,5=0,6mol$

$a. PTHH : $

                    $2Al  +   6HCl\to 2AlCl_3+3H_2↑$

Theo pt :     2 mol        6 mol

Theo đbài :  0,45 mol  0,6 mol

⇒Sau pư Al dư 0,25mol

$m_{Al dư}=0,25.27=6,75g$

c.Theo pt :

$n_{H_2}=1/2.n_{HCl}=1/2.0,6=0,3mol$

$n_{AlCl_3}=1/3.n_{HCl}=1/3.0,6=0,2mol$

$⇒m_{AlCl_3}=0,2.133,5=26,7g$

$m_{ddspu}=12,15-6,75+109,5-0,3.2=114,27g$

$⇒C\%_{AlCl_3}=\dfrac{26,7}{114,27}.100\%=23,37\%$