a. PTHH : Fe2O3 + 6HCl -> 2FeCl3 + 3H2O
a 2a
PTHH : MgO + 2HCl -> MgCl2 + H2O
b b
b. Gọi \(\left\{{}\begin{matrix}n_{Fe_2O_3}=a\left(mol\right)\\n_{MgO}=b\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow160a+40b=24\left(g\right)\left(1\right)\)
\(\Rightarrow m_{muối}=2a.162,5+95b=51,5\left(g\right)\left(2\right)\)
Từ (1) (2 ) => a = 0,1 ; b = 0,2 mol
\(n_{HCl}=0,6+0,4=1\left(mol\right)\)
\(\Rightarrow m_{HCl}=\dfrac{1.36,5}{14,6\%}=250\left(g\right)\)
\(\%m_{Fe_2O_3}=\dfrac{0,1.160}{24}.100=66,66\%\)
\(\%m_{MgO}=100\%-66,66\%=33,34\%\)