Zn+2HCl->ZnCl2+H2
0,1..0,2.....0,1.........0,1
nZn=6,5/65=0,1mol
mHCl=7,3.125/100=9,125g=>nHCl=0,25mol
=>Zn hết
VH2(₫ktc)=0,1.22,4=2,24(l)
mZnCl2=136.0,1=13,6g
mdd sau pứ=6,5+125-0,1.2=131,3g
C% dd ZnCl2....
mHCl dư=(0,25-0,2).36,5=1,825(gl
C% dd HCl dư....