* m X = 6,59 (g)
m muối = 14,185 (g)
V HCl = 100 (ml) = 0,1 (l)
Cm HCl = 2,5 (M)
=> n HCl = Cm . V = 2,5 . 0,1 = 0,25 (mol)
* PTHH
Fe + 2 HCl --> FeCl2 + H2
a------------2a----------a-------a
Fe3O4 + 8 HCl --> FeCl2 + 2 FeCl3 + 4 H2O
b------------8b---------b--------2b---------4b
2 Al + 6 HCl --> 2 AlCl3 + 3 H2
2c-----------6c----------2c--------3c
* Ta có hệ 3 pt :
56 . a + 232 . b + 27 . 2c = 6,59
2a + 8b + 6c = 0,25
127 . (a + b) + 162,5 . 2b + 133,5 . 2c = 14,185
<=> a = 0,03
b = 0,02
c = 0,005
* m Fe = 56 . a = 56 . 0,03 = 1,68
=> %m Fe = 1,68/6,59 . 100 % = 25,5%
* n H2 = a + 3c = 0,045 (mol)
=> V H2 = 0,045 . 22,4 = 1,008 (l)