Ta có FeSO4(a mol) MgSO4(b mol) K2SO4( c mol)Ta có FeSO4(a mol) MgSO4
(b mol) K2SO4( c mol)
a+b+c=4ba+b+c=4b
⇒a-3b+c=0⇒a-3b+c=0
⇒{152a+120b+174c=88,05127a+95b+149c=73,05
⇒a=0,375b=0,15 c=0,075⇒{152a+120b+174c=88,05127a+95b+149c=73,05
⇒{a=0,375b=0,15 c=0,075
VBaCl2=0,6/2=0,3(l) VBaCl2=0,6/2=0,3(l)
⇒mBaSO4=0,6x233=139,8(g)⇒mBaSO4=0,6x233=139,8(g)
b, ⇒mFeSO4=57(g)mMgSO4=18(g)mK2SO4=13,05(g)
⇒{mFeSO4=57(g)mMgSO4=18(g)mK2SO4=13,05(g)
c,nKOH=0,9(mol)nKOH=0,9(mol)
PTHH:FeCl2+2KOH→Fe(OH)2+2KClPTHH:FeCl2+2KOH→Fe(OH)2+2KCl
4Fe(OH)2+O2+2H2O→4Fe(OH)34Fe(OH)2+O2+2H2O→4Fe(OH)3
2Fe(OH)3→Fe2O3+3H2O2Fe(OH)3→Fe2O3+3H2O
MgCl2+2KOH→Mg(OH)2+KClMgCl2+2KOH→Mg(OH)2+KCl
Mg(OH)2→MgO+H2OMg(OH)2→MgO+H2O
=>FeCl2 dư
m=0,15x40+0,15x160=30(g)