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$\rm n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$

$\rm PTHH: 2Al+6HCl \to 2AlCl_3+3H_2$

$\rm Al_2O_3+6HCl \to 2AlCl_3+3H_2O$

$\rm a.n_{Al}=\dfrac{2}{3}n_{H_2}=0,2(mol)$

$\rm m_{Al}=0,2.27=5,4(g)$

$\rm \%Al=\dfrac{5,4}{36}.100\%=15\%$

$\rm \% Al_2O_3=100-15=75\%$

$\rm b.n_{Al_2O_3}=\dfrac{36-5,4}{102}=0,3(mol)$

$\rm n_{HCl}=3n_{Al}+6n_{Al_2O_3}=0,2.3+0,3.6=2,4(mol)$

$\rm m_{HCl}=2,4.36,5=87,6(g)$

$\rm m_{dd HCl}=\dfrac{87,6.100}{14,6}=600(g)$

`=>`$\rm V=\dfrac{600}{1,2}=500(ml)$

$\rm c.n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=0,2+0,3.2=0,8(mol)$

$\rm m_{dd AlCl_3}=36+600-0,3.2=635,4(g)$

$\rm C\% AlCl_3=\dfrac{0,8.133,5.100\%}{635,4}=16,81\%$