a)
PTHH:
$2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2$

$n_{Al}=\dfrac{5,4}{27}=0,2$ (mol)
$m_{H_2SO_4}=400\cdot 9,8%=39,2$ (g)
$\Rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4$ (mol)

$0,2$ mol $Al$ cần $0,3$ mol $H_2SO_4$
Mà $0,4>0,3$ ⇒ $H_2SO_4$ dư

$n_{H_2SO_4,\text{dư}}=0,4-0,3=0,1$ (mol)
$\Rightarrow m_{\text{dư}}=0,1\cdot 98=9,8$ (g)

b)
$n_{H_2}=\dfrac{3}{2}\cdot 0,2=0,3$ (mol)
$\Rightarrow V_{H_2}=0,3\cdot 22,4=6,72$ (lít)

$n_{Al_2(SO_4)3}=\dfrac{0,2}{2}=0,1$ (mol)
$\Rightarrow m{\text{muối}}=0,1\cdot 342=34,2$ (g)

$m_{\text{dd}}=400+5,4-0,3\cdot 2=404,8$ (g)

$C_{\text{% }Al_2(SO_4)_3}=\dfrac{34,2}{404,8}\cdot 100\approx 8,45%$

$C_{\text{% }H_2SO_4,\text{dư}}=\dfrac{9,8}{404,8}\cdot 100\approx 2,42%$