a,
$n_{Al}=\dfrac{5,4}{27}=0,2(mol)$
$n_{H_2SO_4}=\dfrac{200.15\%}{98}=\dfrac{15}{49}(mol)\approx 0,306(mol)$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$
$\to Al$ hết, axit dư
$\to n_{H_2}=\dfrac{3}{2}n_{Al}=0,3(mol)$
Vậy $V_{H_2}=0,3.22,4=6,72l$
b,
$m_{\text{dd spu}}=5,4+200-0,3.2=204,8g$
$n_{Al_2(SO_4)_3}=0,5n_{Al}=0,1(mol)$
$\to C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342.100}{204,8}=16,7\%$
$n_{H_2SO_4\rm dư}=\dfrac{15}{49}-0,3=0,006(mol)$
$\to C\%_{H_2SO_4}=\dfrac{0,006.98.100}{204,8}=0,287\%$