$n_{AlCl_3}=0,1(mol)$
$n_{Al_2O_3}=\dfrac{2,55}{102}=0,025(mol)$
$2Al(OH)_3\buildrel{{t^o}}\over\to Al_2O_3+3H_2O$
$\Rightarrow n_{Al(OH)_3}=0,025.2=0,05(mol)$
- TH1: tủa chưa tan
$AlCl_3+3NaOH\to Al(OH)_3+3NaCl$
$\Rightarrow n_{NaOH}=0,05.3=0,15(mol)$
$\to C_{M_{NaOH}}=\dfrac{0,15}{0,2}=0,75M$
- TH2: tủa tan
$AlCl_3+3NaOH\to Al(OH)_3+3NaCl$ (1)
$n_{NaOH(1)}=0,1.3=0,3(mol)$
$n_{Al(OH)_3(2)}=0,1-0,05=0,05(mol)$
$Al(OH)_3+NaOH\to NaAlO_2+2H_2O$ (2)
$n_{NaOH(2)}=0,05(mol)$
$\to C_{M_{NaOH}}=\dfrac{0,3+0,05}{0,2}=1,75M$
