Đáp án:
nH2=$\frac{11,2}{22,4}$ =0,5
27nAl+65nZn=18,4
3nAl+2nZn=2nH2=0,5.2=1
=>$\left \{ {{nAl=0,2} \atop {nZn=0,2}} \right.$
%mAl=$\frac{0,2.27}{18,4}$.100=29,35%
%mZn=100-29,35=70,65%
nHCl=2nH2=0,5.2=1
VddHCl=$\frac{1}{2}$=0,5lit
CMAlCl3=CM ZnCl2=$\frac{0,2}{0,5}$=0,4M