$m_{dd\ sau\ pư} = 9,18 + 98,55 = 107,73(gam)$
$\Rightarrow m_{muối} = 107,73.22,306\% = 24,03(gam)$
$R_2O_3 + 6HCl \to 2RCl_3 + 3H_2O$
Theo PTHH :
$2n_{R_2O_3} = n_{RCl_3}$
$\Rightarrow \dfrac{2.9,18}{2R + 16.3} = \dfrac{24,03}{R + 35,5.3}$
$\Rightarrow R = 27(Al)$
$C\% = \dfrac{0,09.6.36,5}{98,55}.100\% = 20\%$
Tên : Nhôm oxit