a,
$n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)$
$Fe+2HCl\to FeCl_2+H_2$
$\to n_{Fe}=n_{FeCl_2}=n_{H_2}=0,15(mol)$
$\to n_{MgO}=\dfrac{12,4-0,15.56}{40}=0,1(mol)$
$MgO+2HCl\to MgCl_2+H_2O$
$\to n_{MgCl_2}=n_{Mg}=0,1(mol)$
$m_{\text{muối}}=0,1.95+0,15.127=28,55g$
b,
$n_{HCl\text{pứ}}=2n_{Fe}+2n_{MgO}=0,15.2+0,1.2=0,5(mol)$
$\to n_{HCl\text{dư}}=0,5.10\%=0,05(mol)$
$NaOH+HCl\to NaCl+H_2O$
$2NaOH+FeCl_2\to Fe(OH)_2+2NaCl$
$2NaOH+MgCl_2\to Mg(OH)_2+2NaCl$
$n_{NaOH}=0,05+0,15.2+0,1.2=0,55(mol)$
$\to V=0,55l$