$2Na+2HCl\to 2NaCl+H_2\ (1)\\ 2Na+2H_2O\to 2NaOH+H_2\ (2)\\ n_{HCl(bd)}=\dfrac{500}{1000}.0,2=0,1(mol)\\ NaOH+HCl\to NaCl+H_2O\ (3)\\ n_{HCl(3)}=\dfrac{100}{1000}.0,2=0,02(mol)\\ n_{NaOH(3)}=n_{HCl(3)}=0,02(mol)\\ \sum n_{Na}=n_{HCl(bd)}+n_{NaOH(3)}=0,12(mol)\\ m_{Na}=0,12.23=2,76(g)\\ \sum n_{H_2}=\dfrac{1}{2}n_{Na}=0,06(mol)\\ V_{H_2}=0,06.22,4=1,344(l)$

Đáp án:

$\rm n_{HCl\ (1)}=500×10^{-3}×0,2=0,1\ (mol)\\n_{HCl\ (2)}=100×10^{-3}×0,2=0,02\ (mol)$

$\rm\sum n_{Cl^-}=0,1+0,02=0,12\ (mol)$

Bảo toàn điện tích:

$\rm n_{Na^+}=\sum n_{Cl^-}=0,12\ (mol)$

$\rm m_{Na}=0,12×23=2,76\ (g)$

Bảo toàn electron:

$\rm n_{H_2}=\dfrac{1}{2}n_{Na}=0,06\ (mol)$

$\rm V_{H_2}=0,06×22,4=1,344\ (L)$