Giải thích các bước giải:
nNaOH=0,6.2=1,2(M)
FeCl3+3NaOH---->Fe(OH)3+3NaCl
0,4 1,2 0,4 1,2
a)Theo pt:nFeCl3=1/3.nNaOH=1,2/3=0,4(mol)
=>VFeCl3=0,4/0,2=2(l)
b)Theo pt:nFe(OH)3=nFeCl3=0,4(mol)
=>mFe(OH)3=0,4.107=42,8(g)
c)Theo pt:nNaCl=nNaOH=1,2(mol)
=>CmNaCl=$\frac{1,2}{2+0,6}$ =0,4615(M)
nNaOH=0,6.2=1,2mol
FeCl3+3NaOH→Fe(OH)3+3NaCl
nFeCl3=1,2/3=0,4mol
Vdd FeCl3=0,4/0,2=2l
mFe(OH)3=0,4.107=42,8g
Vdd sau pứ=0,6+2=2,6l
nNaCl=nNaOH=1,2mol
CM NaCl=1,2/2,6=0,46M
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