a) CaO+2HCl---->CaCl2+H2O
x--------2x
Fe2O3+6HCl---->2FeCl3+3H2O
y------6y
n HCl=0,2.3,3=0,66(mol)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}56x+160y=17,76\\2x+6y=0,66\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,06\\y=0,09\end{matrix}\right.\)
%m CaO=0,06.56/17,76.100%=18,9%
%m Fe2O3=100%-18,9%=81,1%