FeO+2HCl−−>FeCl2+H2OFeO+2HCl−−>FeCl2+H2O
Fe2O3+6HCl−−>2FeCl3+3H2OFe2O3+6HCl−−>2FeCl3+3H2O
theo đề ta có nFeO=nFe2O3nFeO=nFe2O3
=> nFeO=nFe2O3=11.6/(72+160)=0.05molnFeO=nFe2O3=11.6/(72+160)=0.05mol
nHCl=2∗0.3=0.6molnHCl=2∗0.3=0.6mol
theo PT : nHCl=0.4molnHCl=0.4mol
=> nHCldư=0.2molnHCldư=0.2mol
nFeCl2=nFeO=0.05molnFeCl2=nFeO=0.05mol
nFeCl3=2nFe2O3=0.1molnFeCl3=2nFe2O3=0.1mol
=> CM FeCl2,FeCl3,HCldưFeCl2,FeCl3,HCldư =.....................
b. gợi ý: trong dung dịch A có: HCldư,FeCl2,FeCl3HCldư,FeCl2,FeCl3
HCl+NaOH−−>NaCl+H2OHCl+NaOH−−>NaCl+H2O
FeCl2+2NaOH−−>Fe(OH)2+2NaClFeCl2+2NaOH−−>Fe(OH)2+2NaCl
FeCl3+3NaOH−−>Fe(OH)3+3NaClFeCl3+3NaOH−−>Fe(OH)3+3NaCl
từ PT=> nNaOH