Привести реакции взаимодействия избытка NaOH c: CrCl3, FeSO4, AlCl3

CrCl3 + 6NaOH = Na3[Cr (OH) 6] + 3NaCl

FeSO4 + 2NaOH = Fe (OH) 2↓ + Na2SO4

AlCl3 + 4NaOH = Na[Al (OH) 4] + 3NaCl

3NaOH + CrCL3 = Cr (OH) 3 + 3NaCL

Cr (3+) + 3OH - = Cr (OH) 3

Cr (OH) 3 + NaOH = NaCrO2 + 2H2O

Cr (OH) 3 + OH-=CrO2 - + H2O

FeSO4 + 2NaOH = Fe (OH) 2 + Na2SO4

Fe (2+) + 2OH-=Fe (OH) 2

AlCL3 + 3NaOH = Al (OH) 3 + 3 NaCL

Al (3+) + 3OH - = AL (OH) 3

AL (OH) 3 + NaOH = NaALO2 + H2O

Al (OH) 3 + OH - = ALO2 - + 2H2O