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Gọi số mol của FeO và Fe2O3 là a (mol)
Theo đề bài, ta có: \(m_{hh}=11,6\left(g\right)\)
\(\Rightarrow72a+160a=11,2\)
\(\Rightarrow a=0,05\left(mol\right)\)PTHH: \(FeO+2HCl\rightarrow FeCl_2+H_2O\)
pư............0,05...........0,1..............0,05..............0,05 (mol)
PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
pư.............0,05...........0,3.............0,1...............0,15 (mol)
b) \(\Rightarrow\left\{{}\begin{matrix}m_{FeO}=72.0,05=3,6\left(g\right)\\m_{Fe2O3}=11,6-3,6=8\left(g\right)\end{matrix}\right.\)
c) Ta có: \(n_{HCl\left(dư\right)}=0,2.3-\left(0,1+0,3\right)=0,2\left(mol\right)\)
\(\left\{{}\begin{matrix}C_{M\left(FeCl2\right)}=\dfrac{0,05}{0,2}=0,25\left(M\right)\\C_{MFeCl3}=\dfrac{0,1}{0,2}=0,5\left(M\right)\\C_{M\left(HCldư\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)
Vậy..............