$R_2O_n+2nHCl\to 2RCl_n+nH_2O$
$n_{R_2O_n}=\dfrac{2,04}{2R+16n}$
$\Rightarrow n_{RCl_n}=\dfrac{4,08}{2R+16n}$
$\Rightarrow \dfrac{4,08}{2R+16n}=\dfrac{5,34}{R+35,5n}$
$\Leftrightarrow 5,34(2R+16n)=4,08(R+35,5n)$
$\Leftrightarrow R=9n$
$n=3\Rightarrow R=27(Al)$
$\to Al_2O_3$