$m_{H_2SO_4}=9,8\%.300=29,4g$
$⇒n_{H_2SO_4}=\dfrac{29,4}{98}=0,3mol$
$n_{R_2O_3}=\dfrac{10,2}{2.M_R+48}(mol)$
$PTPU :$
$R_2O_3+3H_2SO_4\to R_2(SO_4)_3+3H_2O$
$\text{Theo pt :}$
$n_{R_2O_3}=\dfrac{1}{3}.n_{H_2SO_4}=\dfrac{1}{3}.0,3=0,1mol$
$⇒\dfrac{10,2}{2.M_R+48}=0,1$
$⇔M_R=27$
$\text{⇒R là Al}$
$\text{⇒CTHH của oxit là Al2O3}$
$PTHH :$
$Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O$
$\text{Theo pt :}$
$n_{Al_2(SO_4)_3}=n_{Al_2O_3}=\dfrac{10,2}{102}=0,1mol$
$⇒m_{Al_2(SO_4)_3}=0,1.342=34,2g$
$m_{dd\ spu}=10,2+300=310,2g$
$⇒C\%_{Al_2(SO_4)_3}=\dfrac{34,2}{310,2}.100\%=11,03\%$
