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
\(Nxét:0,2< 0,3\Rightarrow NaAlO_2dư\left(t/m\right)\)
Ta có: \(t+0,2=1\Rightarrow t=0,8\left(t/m\right)\)
\(\rightarrow m_{NaOH}=0,8.40=32\left(g\right)\)
TH2: Nếu kết tủa không bị tan
\(NaOH+HCl\rightarrow NaCl+H_2O\)
t------------>t
\(NaAlO_2+HCl+H_2O\rightarrow NaCl+Al\left(OH\right)_3\downarrow\)
0,3---------->0,3-------------------------->0,3
\(n_{Al\left(OH\right)_3\left(tan\right)}=0,3-0,2=0,1\left(mol\right)\)
\(Al\left(OH\right)_3+3HCl\rightarrow AlCl_3+3H_2O\)
0,1--------->0,3
\(\Rightarrow t+0,3+0,3=1\Rightarrow t=0,4\left(t/m\right)\\ \Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\)