a,
$n_{Fe}=\dfrac{22,4}{56}=0,4(mol)$
$n_{H_2SO_4}=\dfrac{24,5}{98}=0,25(mol)$
$Fe+H_2SO_4\to FeSO_4+H_2$
$\dfrac{0,4}{1}>\dfrac{0,25}{1}$
$\Rightarrow Fe$ dư
$n_{Fe\text{pứ}}=n_{H_2SO_4}=0,25(mol)$
$\Rightarrow n_{Fe\text{dư}}=0,4-0,25=0,15(mol)$
$\to m_{\text{dư}}=0,4.56=8,4g$
b,
$n_{H_2}=n_{H_2SO_4}=0,25(mol)$
$\to V_{H_2}=0,25.22,4=5,6l$