\(n_{Al_2O_3}=\dfrac{8.1}{102}=\dfrac{27}{340}\left(mol\right)\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
\(\dfrac{27}{340}.......\dfrac{81}{170}\)
\(m_{HCl}=\dfrac{81}{170}\cdot36.5=17.4\left(g\right)\)
\(m_{dd_{HCl}}=\dfrac{17.4}{7.3\%}=238.3\left(g\right)\)
