$n_{H_2}=\dfrac{1,792}{22,4}=0,08 mol$

$Fe+2HCl\to FeCl_2+H_2$

$Zn+2HCl\to ZnCl_2+H_2$

$\Rightarrow n_{HCl\text{pứ}}=2n_{H_2}=0,16 mol$

$n_{HCl}=0,5.0,4=0,2 mol$

$\Rightarrow n_{HCl\text{dư}}=0,2-0,16=0,04 mol$

$NaOH+HCl\to NaCl+H_2O$

$\Rightarrow n_{NaOH}=0,04 mol$

$\to V_{NaOH}=\dfrac{0,04}{0,5}=0,08l=80ml$

Đáp án:

\(\begin{array}{l} V_{NaOH}=0,08\ lít.\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} PTHH:\\ Fe+2HCl\to FeCl_2+H_2↑\ (1)\\ Zn+2HCl\to ZnCl_2+H_2↑\ (2)\\ NaOH+HCl\to NaCl+H_2O\ (3)\\ n_{HCl}=0,5\times 0,4=0,2\ mol.\\ n_{H_2}=\dfrac{1,792}{22,4}=0,08\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,2}{2}>\dfrac{0,08}{1}\\ ⇒HCl\ \text{dư.}\\ ⇒n_{HCl}\text{(dư)}=0,2-(2\times 0,08)=0,04\ mol.\\ Theo\ pt\ (3):\ n_{NaOH}=n_{HCl}=0,04\ mol.\\ ⇒V_{NaOH}=\dfrac{0,04}{0,5}=0,08\ lít.\end{array}\)

chúc bạn học tốt!