Đáp án:

$a,\ M:\ Al\\ X:\ Al_2(SO_4)_3.18H_2O\\ b,\ V_{dd\ NaOH}=1200(ml)$

Giải thích các bước giải:

$a,\\ X:\ M_2(SO_4)_n.aH_2O\\ P2:\\ n_{BaSO_4}=\dfrac{13,98}{233}=0,06(mol)\\ M_2(SO_4)_n+nBaCl_2\to nBaSO_4+2MCl_n\\ \xrightarrow{\text{Từ PTHH}}\ n_{M_2(SO_4)_n}=\dfrac{1}{n}.n_{BaSO_4}=\dfrac{0,06}{n}(mol)\\ P1:\\ M_2(SO_4)_n+2nNH_3+2nH_2O\to 2M(OH)_n\downarrow+n(NH_4)_2SO_4\\ 2M(OH)_n\xrightarrow{t^0} M_2O_n+nH_2O\\ \xrightarrow{\text{Từ PTHH}}\ n_{M_2O_n}=n_{M_2(SO_4)_n}=\dfrac{0,06}{n}(mol)\\ \Rightarrow \dfrac{2,04}{2M+16n}=\dfrac{0,06}{n}\\ \Rightarrow M=9n\\ \to n=3;\ M=27\ (Al)\\ \to M:\ Al\\ \Rightarrow n_{Al_2O_3}\dfrac{2,04}{102}=0,02(mol)\\ \Rightarrow \sum\limits n_{X}=2.n_{Al_2O_3}=0,04(mol)\\ \Rightarrow M_X=\dfrac{26,64}{0,04}=666(g/mol)\\ \Rightarrow a=18\\ \to X:\ Al_2(SO_4)_3.18H_2O$

$b,\\ n_A=n_X=0,04(mol)$

Để kết tủa lớn nhất thì phản ứng xảy ra vừa đủ, $Al(OH)3$ không tan

$6NaOH+Al_2(SO_4)_3\to 2Al(OH)_3+3Na_2SO_4\\ \xrightarrow{\text{Từ PTHH}}\ n_{NaOH}=6.n_{Al_2(SO_4)_3}=0,24(mol)\\ \Rightarrow V_{dd\ NaOH}=\dfrac{0,24}{0,2}=1,2(l)=1200(ml)\\ \to V_{dd\ NaOH}=1200(ml)$