28.
nAl(3+) = 0,2x
nSO4(2-) = 0,3x
nOH- = 0,6x
nBa(2+) = 0,3x
2Al(OH)3 -> Al2O3 + 3H2O
m giảm = mAl(OH)3 - mAl2O3 = 0,2x.78 - 0,1x.102 = 5,4
-> x = 1 -> B
30.
Al + NaOH + H2O -> NaAlO2 + 3/2H2
0,4----0,4----------------------------------0,6
Al2O3 + 2NaOH -> 2NaAlO2 + H2O
--0,2---------0,4-----------------------------
nAl2O3 = (31,2 - 0,4.27)/102 = 0,2 mol
-> Tổng số mol NaOH cần dùng nNaOH = 0,8 mol
-> V = n/CM = 0,8/4 = 0,2 (l) = 200ml
-> V cần dùng = 210 ml
33.
Na + H2O -> NaOH + 1/2H2
-a-------------------a---------a/2--
NaOH + Al + H2O -> NaAlO2 + 3/2H2
---a--------a-----------------------------3a/2--
-> a/2 + 3a/2 = 0,3 -> a = 0,15
m = (27+23).0,15 = 7,5 (g)