\(n_{BaSO_4}=\dfrac{0.699}{233}=0.003\left(mol\right)\)
\(Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow3BaSO_4+2AlCl_3\)
\(0.001...............................0.001\)
\(M_{tt}=\dfrac{6.66}{0.001}=666\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow342+18n=666\)
\(\Rightarrow n=18\)
\(Al_2\left(SO_4\right)_3\cdot18H_2O\)
$n_{BaSO_4} = 0,003(mol)$
$n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{BaSO_4} = 0,001(mol)$
Trong dd A : $n_{Al_2(SO_4)_3} = 0,001.10 = 0,01(mol)$
Suy ra :
$M_{tinh\ thể} = 342 + 18n = \dfrac{6,66}{0,01} = 666 \Rightarrow n = 18$
Vậy CT tinh thể là $Al_2(SO_4)_3.18H_2O$
