\(n_{Cl_2}=\dfrac{v}{22,4}=\dfrac{14,56}{22,4}=0,65mol\)
-Gọi nCu=x, nAl=y,nFe=z
64x+27y+56z=23,8(I)
Cu+Cl2\(\overset{t^0}{\rightarrow}\)CuCl2
2Al+3Cl2\(\overset{t^0}{\rightarrow}\)2AlCl3
2Fe+3Cl2\(\overset{t^0}{\rightarrow}\)2FeCl3
\(\rightarrow\)\(n_{Cl_2}=n_{Cu}+\dfrac{3}{2}n_{Al}+\dfrac{3}{2}n_{Fe}=x+\dfrac{3}{2}y+\dfrac{3}{2}z=0,65\)(II)
(x+y+z)t=0,25(III)
2Al+6HCl\(\rightarrow\)2AlCl3+3H2
Fe+2HCl\(\rightarrow\)FeCl2+H2
\(\rightarrow\)\(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\left(1.5y+z\right)t=0,2\)(IV)
-Lập tỉ số: \(\dfrac{\left(III\right)}{\left(IV\right)}\)ta có:\(\dfrac{x+y+z}{1,5y+z}=\dfrac{0,25}{0,2}=1,25\)
\(\rightarrow\)x+y+z=1,875y+1,25z\(\rightarrow\)x-0,875y-0,25z=0(V)
-Giải hệ (I,II,V) có được: x=0,2, y=0,2,z=0,1
%Cu=\(\dfrac{0,2.64.100}{23,8}\approx53,78\%\)
%Al=\(\dfrac{0,2.27.100}{23,8}\approx22,7\%\)
%Fe=100%-53,78%-22,7%=23,52%
