47/ nAl(NO3)3 = 0.05
2Al(NO3)3 ---to---> Al2O3
x...............................x/2
=> (0.05-x)213 + 102x/2 = 7.41
=> x = 0.02 => H% = 0,02/0,05.100% = 40%
48/
nH2 = 0.6
Al + NaOH + H2O = NaAlO2 + 3/2H2
0.4...0.4........................................0.6
=> nAl2O3 = 31.2-27*0.4)/102=0.2 mol
Al2O3 + 2NaOH = 2NaAlO2 + H2O
0,2.........0,4
=> nNaOH = 0.8 => Vđem dùng = 0.8/0.5+0.1=1.7l => D
59/
8Al + 3Fe3O4 = 9Fe + 4Al2O3
8/9x.........................x
2Al + 3H2SO4 = Al2(SO4)3 + 3H2
Fe + H2SO4 = FeSO4 + H2
nH2 = 3/2nAl + nFe => 3/2 (0.4 - 8/9x) + x = 0.48 => x=0.36
vì nAl/8 = nFe3O4/3 nên H% tính theo cái nào cũng được = (0.36.8/9)/0.4*100=80%
61/
nkhis = nH2 = 0.25 mol
Na + H2O = NaOH + 1/2 H2
0,5..................0,5.........0,25
nran = nAl2O3 = 0.05 => nAl(OH)3 kt = 0.1 mol
vì thu kt nên NaOH pư hết
3NaOH + AlCl3 = Al(OH)3 + 3NaCl
3x..............x...........x
NaOH + Al(OH)3 = NaAlO2 + 2H2O
0,5-3x......0,5-3x
=> x-(0,5 - 3x) = 0,1 => x = 0.15
=> CM = 0.15/0.4=0.375M