$n_{HCl}=0,2(mol)$
$n_{FeCl_3}=0,2.0,5=0,1(mol)$
$NaOH+HCl\to NaCl+H_2O$
$3NaOH+FeCl_3\to Fe(OH)_3+3NaCl$
$\to n_{NaOH}=n_{HCl}+3n_{FeCl_3}=0,5(mol)$
$\to V=\dfrac{0,5}{1}=0,5M$
Muối trong $Z$ gồm: $NaCl$ ($0,2+0,1.3=0,5$ mol)
$\to m_{NaCl}=0,5.58,5=29,25g$