FeCl2 + 2NaOH -> Fe(OH)2 + 2NaCl (1)
Đặt Vdd NaOH 20%=200(g)
nNaOH=1(mol)
Từ 1:
nFe(OH)2=nFeCl2=\(\dfrac{1}{2}\)nNaOH=0,5(mol)
nNaCl=nNaOH=1(mol)
mdd FeCl2=\(\dfrac{0,5.127}{12,7\%}=500\left(g\right)\)
mFe(OH)2=90.0,5=45(g)
C% dd NaCl=\(\dfrac{58,5}{200+500-45}.100\%=8,93\%\)