\(n_{Al\left(OH\right)_3}=\dfrac{1,17}{78}=0,015\left(mol\right)\\ n_{AlCl_3}=\dfrac{2,67}{133,5}=0,02\left(mol\right)\)
TH1: Kết tủa không bị tan
\(NaOH+HCl\rightarrow NaCl+H_2O\)
a<------------a
\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\)
0,015<--0,045<------0,015
\(Nxét:0,015< 0,02\Rightarrow AlCl_3dư\left(t/m\right)\)
Ta có: \(a+0,045=0,5\Rightarrow a=0,455\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,455.40=18,2\left(g\right)\)
TH2: Kết tủa tan một phần
\(NaOH+HCl\rightarrow NaCl+H_2O\)
a<------------a
\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\)
0,02---->0,06-------->0,015
\(n_{Al\left(OH\right)_3\left(tan\right)}=0,02-0,015=0,005\left(mol\right)\)
\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)
0,005------>0,005
\(\Rightarrow a+0,06+0,005=0,5\Rightarrow a=0,435\left(mol\right)\\ \Rightarrow m_{NaOH}=0,435.40=17,4\left(g\right)\)
