$n_{HCl}=2.0,05=0,1mol$
$n_{M_2O_3}=\dfrac{2,4}{2.M_M+48}(mol)$
$\text{Gọi oxit của kim loại là}$ $M_2O_3$
$PTPU :$
$M_2O_3+6HCl\to 2MCl_3+3H_2O$
$\text{Theo pt :}$
$n_{M_2O_3}=\dfrac{1}{6}.n_{HCl}=\dfrac{1}{6}.0,1=\dfrac{1}{60}mol$
$⇒\dfrac{2,4}{2.M_M+48}=\dfrac{1}{60}$
$⇔144=2M+48$
$⇒M=48(???)$