Question

Answer 1

\(Đặt:n_{CuO}=a\left(mol\right);n_{Fe_2O_3}=b\left(mol\right)\left(a,b>0\right)\\ a.PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ n_{HCl}=0,2.3,5=0,7\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}80a+160b=20\\2a+6b=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\\ m_{CuO}=80a=80.0,05=4\left(g\right)\\ \Rightarrow m_{Fe_2O_3}=20-4=16\left(g\right)\)