3Fe3O4 + 28HNO3 --> 9Fe(NO3)3 + NO + 14H2O
x------------------------3x
nNO3- = 9x
3Cu + 8H+ + 2NO3- --> 3Cu2+ + 2NO + 4H2O
13,5x---------9x
Cu + 2Fe3+ --> Cu2+ + 2Fe2+
1,5x--3x
13,5x + 1,5x = 0,27 => x = 0,018
=> a = 4,176g
Na --> Na+ + e
a------------a
Mg --> Mg2+ + 2e
b--------------2b
Al --> Al3+ + 3e
c-------------3c
N+5 + 4e --> N+1
------4x------x
a + 2b + 3c = 4x
a + 2b + 3c + x = 3,3
=> x = 0,66 => V = 0,66/2.22,4 = 7,392 lít
nNO3- trong muối = a + 2b + 3c = 2,64 => m muối = 7,92 + 2,64.62 = 48,84g
