Zn+ 2HCl\(\rightarrow\)ZnCl2+ H2
Fe+ 2HCl\(\rightarrow\)FeCl2+ H2
nH2= \(\frac{6,72}{22,4}\)= 0,3 mol
Đặt nZn= x mol; nFe= y mol \(\left\{{}\begin{matrix}\text{65x+56y= 18,6}\\\text{x+y=0,3 }\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}\text{x=0,2}\\\text{y=0,1}\end{matrix}\right.\)
BTNT, nZn= nZnCl2= 0,2 mol
nFe= nFeCl2= 0,1 mol
\(\rightarrow\) mZnCl2= 0,2.136= 27,2g
mFeCl2= 0,1.127= 12,7g
\(\rightarrow\)%ZnCl2= \(\frac{27,2.100}{\text{27,2+12,7}}\)= 68,17%
\(\rightarrow\) %FeCl2= 31,83%
