$a.n_{M_2O_3}=\dfrac{10,2}{2.M_M+48}(mol)$
$m_{HCl}=20\%.109,5=21,9g$
$⇒n_{HCl}=\dfrac{21,9}{36,5}=0,6mol$
Gọi oxit kim loại là $M_2O_3$
$PTPU :$
$M_2O_3+6HCl\to 2MCl_3+3H_2O$
Theo pt :
$n_{M_2O_3}=\dfrac{1}{6}.n_{HCl}=\dfrac{1}{6}.0,6=0,1mol$
$⇒\dfrac{10,2}{2.M_M+48}=0,1$
$⇒2.M_M+48=102$
4⇔2.M_M=54$
$⇔M_M=27(Al)$
⇒M là nhôm ⇒ Oxit kim loạilà Al2O3
$b.PTHH :$
$Al_2O_3+6HCl\to 2AlCl_3+3H_2O$
Theo pt :
$n_{AlCl_3}=\dfrac{1}{3}.n_{HCl}=\dfrac{1}{3}.0,6=0,2mol$
$⇒m_{AlCl_3}=0,2.133,5=26,7g$
$m_{dd\ spu}=10,2+109,5=119,7g$
$⇒C\%_{AlCl_3}=\dfrac{26,7}{119,7}.100\%=22,3\%$