b) Theo pthh1
n\(_{FeCl2}=n_{Fe}=0,2\left(mol\right)\)
mdd=200+20-0,6=219,4(g)
C% FeCl3=\(\frac{0,2.127}{219,4}.100\%=11,58\%\)
Theo pthh2
n\(_{ZnCl2}=n_{Zn}=0,1\left(mol\right)\)
C%ZnCl2=\(\frac{0,1.136}{119,4}.100\%=11,39\%\)
C% HCl dư=\(\frac{0,4.36,5}{219,4}.100\%=6,65\%\)