Câu 6:
TH1: Al(OH)3 không tan
PTHH: 3NaOH+AlCl3→Al(OH)3↓+3NaCl
0,33 0,11 0,11
nAl(OH)3=$\frac{8,58}{78}$=0,11(mol)
=>nAlCl3 pư=0,11<0,15
=>NaOH hết,AlCl3 dư
Theo pt=>nNaOH=3.nAl(OH)3=0,33 (mol)
=>Vdd NaOH=$\frac{0,33}{1}$=0,33 (l)=330ml
TH2: Al(OH)3 tan ít
PTHH: 3NaOH+AlCl3→Al(OH)3+3NaCl
Al(OH)3+NaOH→NaAlO2+2H2O
nNaOH dư=0,15-0,11=0,04 (mol)
=>nNaOH=3.nAlCl3+nNaOH dư=0,49(mol)
=>Vdd NaOH=0,49(l)=490ml
