$n_{Al}=\dfrac{2,7}{27}=0,1(mol)$
$n_{H_2SO_4}=\dfrac{29,4}{98}=0,3(mol)$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$
$\dfrac{0,3}{3}>\dfrac{0,1}{2}$
$\Rightarrow H_2SO_4$ dư
$n_{H_2}=n_{H_2SO_4\text{pứ}}=1,5n_{Al}=0,15(mol)$
$\Rightarrow n_{H_2SO_4\text{dư}}=0,3-0,15=0,15(mol)$
$m_{H_2SO_4\text{dư}}=0,15.98=14,7g$
$V_{H_2}=0,15.22,4=3,36l$