$n_{H_2}=\dfrac{1,792}{22,4}=0,08 mol$
$Fe+2HCl\to FeCl_2+H_2$
$Zn+2HCl\to ZnCl_2+H_2$
$\Rightarrow n_{HCl\text{pứ}}=2n_{H_2}=0,16 mol$
$n_{HCl}=0,5.0,4=0,2 mol$
$\Rightarrow n_{HCl\text{dư}}=0,2-0,16=0,04 mol$
$NaOH+HCl\to NaCl+H_2O$
$\Rightarrow n_{NaOH}=0,04 mol$
$\to V_{NaOH}=\dfrac{0,04}{0,5}=0,08l=80ml$