a, \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Gọi a là số mol của Fe2O3 , b là số mol của CuO
\(\Rightarrow\frac{160a.100}{160a+80b}=80\)
\(\Rightarrow16000a=12800a+6400b\Rightarrow a-2b=0\left(1\right)\)
\(n_{HCl}=0,14\left(mol\right)\Rightarrow6a+2b=0,14\left(2\right)\)
\(\left(1\right)+\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=0,02\\b=0,01\end{matrix}\right.\)
\(\Rightarrow m=m_{Fe2O3}+m_{CuO}=4\left(g\right)\)
b,
\(m_{FeCl3}=162,5.2a=6,5\left(g\right)\)
\(m_{CuCl2}=135.b=1,35\left(g\right)\)
\(V_{dd_{HCl}}=140\left(ml\right)\Rightarrow m_{dd}=140.1,2=168\left(g\right)\)
\(\Rightarrow m_{dd_{spu}}=168+4=172\left(g\right)\)
\(C\%_{FeCl3}=\frac{6,5.100}{172}=3,78\%\)
\(C\%_{CuCl2}=\frac{1,35.100}{172}=0,78\%\)
