\(\begin{array}{l}
n_{KAl(SO_4)_2.12H_2O}=\frac{47,4}{474}=0,1(mol)\\
\to X:\begin{cases} n_{Al^{3+}}=0,1(mol)\\ n_{SO_4^{2-}}=2.0,1=0,2(mol)\end{cases}\\
200ml=0,2l\\
n_{Ba(OH)_2}=0,2.1=0,2(mol)\\
\to \begin{cases} n_{Ba^{2+}}=0,2(mol)\\ n_{OH^-}=0,2.2=0,4(mol)\end{cases}\\
Al^{3+}+3OH^-\to Al(OH)_3(1)\\
Al(OH)_3+OH^-\to AlO_2^-+2H_2O(2)\\
Ba^{2+}+SO_4^{2-}\to BaSO_4(3)\\
n_{OH^-(1)}=3n_{Al^{3+}}=0,3(mol);n_{Al(OH)_3(1))}=n_{Al^{3+}}=0,1(mol)\\
\to n_{OH^-(dư\,sau\,1)}=0,4-0,3=0,1(mol)\\
n_{OH^-(sau\,1)}=n_{Al(OH)_3}\to (2)\,hoàn\,toàn\\
\to Ket\,tua\,spu:\,BaSO_4\\
n_{Ba^{2+}}=n_{SO_4^{2-}}\to Puht\\
\to n_{BaSO_4}=0,2(mol)\to m=46,6(g)
\end{array}\)