250ml=0,25l
\(n_{HCl}=CM.V_{dd}\)=2.0,25=0,5(mol)
gọi x,y lần lượt là số mol của MgOvà \(Fe_2O_3\)
PTHH1:\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
x 2x x x
PTHH2:\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
y 6y 2y 3y
ta có hệ pt:\(\left\{{}\begin{matrix}m_{MgO}+m_{Fe_2O_3}=12\left(g\right)\\n_{HCl\left(1\right)}+n_{HCl\left(2\right)}=0,5\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}40x+160y=12\left(g\right)\\2x+6y=0,5\left(mol\right)\end{matrix}\right.\)
giải ra ta được:x=0,1;y=0,05
\(\%MgO=\dfrac{40.0,1}{12}.100\)=33,33%
\(\%Fe_2O_3=\dfrac{160.0,05}{12}.100\)=66,67%
\(n_{HCl}=0,25.2=0,5\left(mol\right)\)
PTHH: MgO + 2HCl → MgCl2 + H2
Mol: x 2x
PTHH: Fe2O3 + 6HCl → 2FeCl3 + 3H2
Mol: y 3y
Ta có: \(\left\{{}\begin{matrix}40x+160y=12\\2x+3y=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,22\\y=0,02\end{matrix}\right.\)
\(\%m_{MgO}=\dfrac{0,22.40.100\%}{12}=73,33\%\)
\(\%m_{Fe_2O_3}=100-73,33=26,67\%\)
