\(n_{HCl}=0,5.2,2=1,1\left(mol\right)\)
PTHH:
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
x 2x x
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
y 6y 2y
Ta có hệ PT:
\(\left\{{}\begin{matrix}80x+160y=32\\2x+6y=1,1\end{matrix}\right.\)
\(\Rightarrow x=0,1;y=0,15\left(mol\right)\)
\(m_{CuCl_2}=0,1.135=13,5\left(g\right)\)
\(m_{FeCl_3}=0,15.162,5=24,375\left(g\right)\)
\(m_{hh}=13,5+24,375=37,875\left(g\right)\)
\(\%m_{CuCl_2}=\dfrac{13,5}{37,875}.100\%=35,64\%\)
\(\%m_{FeCl_3}=100\%-35,64=64,35\%\)