a,
2NaX $\buildrel{{đpnc}}\over\longrightarrow$ 2Na+ X2
nX2= 0,1 mol => nNaX= 0,2 mol
=> M NaX= 11,7/0,2= 58,5= 23+X
=> X= 35,5 (Cl)
b,
nH2= 0,2 mol
H2+ Cl2 $\buildrel{{askt}}\over\longrightarrow$ 2HCl
=> H2 dư 0,1 mol. Tạo 0,2 mol HCl
=> V HCl= 4,48l
c,
%HCl= $\frac{0,2.100}{0,2+0,1}$= 66,67%
%H2= 33,33%