PTHH:
Al2O3+ 6HCl→ 2AlCl3+ 3H2O
x______ 6x
Fe2O3+ 6HCl→ 2FeCl3+ 3H2O
y _______6y
\(\left\{{}\begin{matrix}102x+160y=34,2\\6x+6y=2\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}x=\frac{287}{870}\\\frac{1}{290}\end{matrix}\right.\)
%Al2O3=\(\frac{\frac{287}{870}.102}{34,2}\text{.100%=98,39%}\)
⇒ %Fe2O3= 100%- 98,39%=1,61%