a) PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
Đặt \(\left\{{}\begin{matrix}n_{Fe_2O_3}=x\left(mol\right)\\n_{MgO}=y\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{FeCl_3}=2x\left(mol\right)\\n_{MgCl_2}=y\left(mol\right)\end{matrix}\right.\)
Ta lập hệ phương trình: \(\left\{{}\begin{matrix}160x+40y=24\\162,5\cdot2x+95y=52,875\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,075\\y=0,3\end{matrix}\right.\)
\(\Rightarrow\%m_{Fe_2O_3}=\dfrac{0,075\cdot160}{24}=50\%=\%m_{MgO}\)
b) Theo PTHH: \(n_{HCl}=6n_{Fe_2O_3}+2n_{MgO}=1,05\left(mol\right)\) \(\Rightarrow C_{M_{HCl}}=a=\dfrac{1,05}{0,5}=2,1\left(l\right)\)
