nNa=0,2 mol, nHCl=0,6 mol
2Na+ 2HCl -> 2NaCl + H2
0,2-->0,2----->0,2---->0,1 (mol)
=> HCl dư 0,6-0,2=0,4 mol
=> VH2=0,1*22,4=...
CNaCl=0,2/0,4=....M
CHCl=0,4/0,4=1M
b. mddHCl=400*1,19=476 g
=>mddsau=mddHCl + mNa -mH2=476+4,6-0,1*2=480,4 gam
C%NaCl=\(\frac{0,2\cdot58,5}{480,4}\cdot100\%=...\)
C%HCl=\(\frac{0,4\cdot36,5}{480,4}\cdot100\%=...\)