a,
$n_{Al}=\dfrac{0,81}{27}=0,03(mol)$
$n_{HCl}=\dfrac{2,29}{36,5}=0,063(mol)$
$2Al+6HCl\to 2AlCl_3+3H_2$
$\dfrac{0,03}{2}>\dfrac{0,063}{6}$
$\Rightarrow Al$ dư
$n_{Al\text{pứ}}=\dfrac{n_{HCl}}{3}=0,021(mol)$
$\Rightarrow n_{Al\text{dư}}=0,03-0,021=0,009(mol)$
$m_{Al\text{dư}}=0,009.27=0,243g$
b,
Sau phản ứng:
$m_{Al}=0,243g$
$n_{AlCl_3}=0,021(mol)$
$\to m_{AlCl_3}=0,021.133,5=2,8035g$
$n_{H_2}=\dfrac{n_{HCl}}{2}=0,0315(mol)$
$\to m_{H_2}=0,0315.2=0,063g$