Gọi \(n_{NaOH}=t\left(mol\right)\)
\(n_{Al\left(OH\right)_3}=\dfrac{15,6}{78}=0,2\left(mol\right)\)
TH1: Nếu kết tủa không bị tan
\(NaOH+HCl\rightarrow NaCl+H_2O\)
t------------>t
\(NaAlO_2+HCl+H_2O\rightarrow NaCl+Al\left(OH\right)_3\downarrow\)
0,2<-------0,2<------------------------0,2
\(Nxét:0,2< 0,25\Rightarrow NaAlO_2dư\left(t/m\right)\)
Ta có: \(t+0,2=0,9\Rightarrow t=0,7\left(t/m\right)\)
\(\rightarrow m_{NaOH}=0,7.40=28\left(g\right)\)TH2: Nếu kết tủa không bị tan
\(NaOH+HCl\rightarrow NaCl+H_2O\)
t------------>t
\(NaAlO_2+HCl+H_2O\rightarrow NaCl+Al\left(OH\right)_3\downarrow\)
0,25---------->0,25-------------------------->0,25
\(n_{Al\left(OH\right)_3\left(tan\right)}=0,25-0,2=0,05\left(mol\right)\)
\(Al\left(OH\right)_3+3HCl\rightarrow AlCl_3+3H_2O\)
0,05--------->0,15
\(\Rightarrow t+0,25+0,15=0,9\Rightarrow t=0,5\left(t/m\right)\\ \Rightarrow m_{NaOH}=0,5.40=20\left(g\right)\)
