a) 
___Zn__+__2HCl--->ZnCl2_+_H2 
___0.04<---0.08<-----0.04<----0.04mol 
m1=0.04*65=2.6g 
mHCl=0.08*36.5=2.92g 
=>m2=m dd HCl 14.6%=2.92*(100/14.6)=20g 
b) 
%chất sau p/ư=%ZnCl2=(mZnCl2/mdd)*100 
mZnCl2=0.04*136=5.44g 
mdd=mZn+mdd HCl-mH2 bay mất 
____=2.6+20-(0.04*2)=22.52g 
=>%ZnCl2=(5.44/22.52)*100=24.16%