Question

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

Answer 1

CrCl3 + 6NaOH = Na3[Cr(OH)6] + 3NaCl
FeSO4 + 2NaOH = Fe(OH)2↓ + Na2SO4
AlCl3 + 4NaOH = Na[Al(OH)4] + 3NaCl

Answer 2

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