Gọi nNaOH = 2a, nBa(OH)2 = a
Pư trung hòa: H+ + OH- -> H2O
=> nOH = 2a + 2a = 4a = 2nH2SO4 = 0.6
=> a = 0.15 mol
=> CM NaOH = 0.15 x 2 /0.3 = 1 M
CM Ba(OH)2 = 0.15 /0.3 = 0.5 M
khi pư Fe2(SO4)3:
Ba2+ + SO42- -> BaSO4 và Fe3+ + 3OH- -> Fe(OH)3
0.15 0.15 0.6 0.2
2Fe(OH)3 -> Fe2O3 + 3H2O
0.2 0.1
=> m chất rắn = mBaSO4 + mFe2O3
= 0.15 x 233 + 0.1 x 160 = 36.55g
MgCl2 -> Mg(OH)2 -> MgO 2FeCl3 -> 2Fe(OH)3 -> Fe2O3
a a b b/2
nOH = 2nMgCl2 + 3nFeCl3 = 2a + 3b = 0.15 mol
m = 3.6 = mMgO + mFe2O3 = 40a + 80b = 3.6
=> a = b = 0.03 mol
Cm MgCl2 = 0.03/0.2 = 0.15 M
Cm FeCl3 = 0.03/0.2 = 0.15 M