Fe3O4 + 8HCl------> 2FeCl3 + FeCl2+4H2O
0,15------1,2----------0,3----------0,15
nFe3O4=34,8/232=0,15
mHCl=455,2.20/100=91,04
nHCl=91,04/36,5=2,5
\(\frac{0,15}{1}< \frac{2,5}{8}\)
mdd=34,8+455,2=490
C%FeCl3=\(\frac{0,3.162,5}{490}.100=9,95\%\)
C%FeCl2=\(\frac{0,15.127}{490}.100=3,89\%\)
C%HCl dư\(\frac{\left(2,5-1,2\right).36,5}{490}100=9,67\%\)
nFe3O4 = 34.8/232=0.15 mol
mHCl = 455.2*20/100=91.04 g
nHCl = 91.04/36.5= 2.5 mol
Fe3O4 + 8HCl --> FeCl2 + 2FeCl3 + 4H2O
Bđ: 0.15______2.5
Pư: 0.15______1.2______0.15____0.3
Kt: 0_________1.3______0.15____0.3
mHCl dư = 47.45 g
mFeCl2 = 19.05 g
mFeCl3 = 48.75 g
mdd A = 34.8 + 455.2 = 490 g
C%HCl dư = 9.68%
C%FeCl2 = 3.88%
C%FeCl3 = 9.95%