Question

Cu ce cantitate de soluție KOH de concentratie 10% ar trebui sa reactioneze 50 grame solutie de FeCl3 de concentrație 25% pentru a obtine un precipitat brun roșcat? Ce cantitate de precipitat se obține? Va rog ajutati-ma cat mai repede!!

Answer 1

MKOH=39+16+1=56g/mol
MFeCl3=56+3*35.5=162.5g/mol
MFe(OH)3=56+3(16+1)=107g/mol

Cp=md/ms*100==>md=Cp*ms/100=25*50/100=12.5g FeCl3

3*56g....162.5g
3KOH+FeCl3➡Fe(OH)3↓+ 3KCl
x=12.923g...12.5g

Cp=md/ms*100==>ms=md*100/Cp=12.923*100/10=129.23g sol KOH

168g KOH......162.5gFeCl3.......107gFe(OH)3
12.923g.........12.5g...................x=107(12.923+12.5)/168+162.5=2720.261/330.5=8.23g precipitat brun-roscat :)

n=m/M=8.23/107=0.0769 moli Fe(OH)3