Vì sau phản ứng có chất rắn không tan nên có oxit dư
Ta có:
\(\left\{{}\begin{matrix}n_{CuO}=0,1\left(mol\right)\\n_{Fe2O3}=0,1\left(mol\right)\\n_{HCl}=0,64\left(mol\right)\end{matrix}\right.\)
Trường hợp 1: CuO tan hết
\(n_{HCl\left(CuO\right)}=2n_{CuO}=0,2\left(mol\right);n_{HCl\left(Fe2O3\right)}=0,64-0,2=0,44\left(mol\right)\)
\(n_{Fe2O3\left(pư\right)}=\frac{0,44}{6}=0,07\left(mol\right)\)
\(n_{Fe2O3\left(dư\right)}=\left(0,1-0,07\right).160=4,8\left(g\right)\)
\(\Rightarrow m=24,375\left(g\right)\Rightarrow m_{muoi}=37,875\left(g\right)\)
Trường hợp 2: Fe2O3 tan hết
\(n_{HCl\left(Fe2O3\right)}=6n_{Fe2O3}=0,6\left(mol\right)\Rightarrow n_{HCl\left(CuO\right)}=0,64-0,6=0,04\left(mol\right)\)
\(n_{CuO\left(Pư\right)}=\frac{1}{2}n_{HCl\left(CuO\right)}=0,02\left(mol\right)\Rightarrow n_{CuO\left(dư\right)}=0,1-0,02=0,08\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeCl3}=32,5\left(g\right)\\m_{CuCl2}=2,7\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{muoi}=35,2\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}4,8< m< 6,4\\35,2< m'< 37,875\end{matrix}\right.\)
