2Na + 2H2O -> 2NaOH + H2 (1)
3NaOH + AlCl3 -> 3NaCl + Al(OH)3 (2)
CT:Al(OH)3 + NaOH -> NaAlO2 + 2H2O (3)
nH2=0,2(Mol)
nAlCl3=0,05(mol)
Theo PTHH 1 ta có:
nNaOH=2nH2=0,4(mol)
Vì 0,05.3<0,4 nên có PƯ 3
Theo PTHH 2 ta có:
nNaCl=3nAlCl3=0,15(mol)
nAl(OH)3=nAlCl3=0,05(mol)
Theo PTHH 3 ta có:
nAl(OH)3=nNaOH(3)=nNaAlO2=0,05(mol)
=>nNaOH còn lại=0,2(mol)
CM dd NaAlO2=\(\dfrac{0,05}{0,5}=0,1M\)
CM dd NaCl=\(\dfrac{0,15}{0,5}=0,3M\)
CM dd NaOH=\(\dfrac{0,2}{0,5}=0,4M\)