a,
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$
b,
$n_{Al_2(SO_4)_3}=\dfrac{68,48}{342}=0,2(mol)$
$\Rightarrow n_{Al}=2n_{Al_2(SO_4)_3}=0,4(mol)$
$m_{Al}=0,4.27=10,8g$
c,
$n_{H_2}=3n_{Al_2(SO_4)_3}=0,6(mol)$
$\Rightarrow V_{H_2}=0,6.22,4=13,44l$
d,
$n_{H_2SO_4}=n_{H_2}=0,6(mol)$
$\Rightarrow C\%_{H_2SO_4}=\dfrac{0,6.98.100}{200}=29,4\%$
e,
$Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3+3Na_2SO_4$
$\Rightarrow n_{Al(OH)_3}=0,2.2=0,4(mol)$
$m_{\downarrow}=0,4.78=31,2g$