\(n_{Al_2\left(SO_4\right)_3}=0,3.0,2=0,06\left(mol\right)\)
\(n_{Al\left(OH\right)_3}=\dfrac{1,56}{78}=0,02\left(mol\right)\)
TH1: NaOH vừa đủ:
\(\dfrac{NaOH}{0,02\left(mol\right)}+\dfrac{Al_2\left(SO_4\right)_3}{0,02\left(mol\right)}\rightarrow\dfrac{Al\left(OH\right)_3\downarrow}{0,02\left(mol\right)}+Na_2SO_4\)
\(\Rightarrow C_{M\left(NaOH\right)}=\dfrac{0,02}{0,5}=0,1\left(M\right)\)
TH2: NaOH dư.
\(\dfrac{NaOH}{0,06\left(mol\right)}+\dfrac{Al_2\left(SO_4\right)_3}{0,06\left(mol\right)}\rightarrow\dfrac{Al\left(OH\right)_3\downarrow}{0,06\left(mol\right)}+Na_2SO_4\)(1)
\(\dfrac{NaOH}{0,02\left(mol\right)}+\dfrac{Al\left(OH\right)_3}{0,02\left(mol\right)}\rightarrow NaAlO_2+2H_2O\)(2)
Từ (1) (2), Suy ra: \(n_{NaOH}=0,06+0,02=0,08\left(mol\right)\)
\(C_{M\left(NaOH\right)}=\dfrac{0,08}{0,5}=0,16\left(M\right)\)