5 cách giải đây nè:
C1: nNO =0.1 mol
Qui đổi Fe,FeO,Fe2O3,Fe3O4 thành Fe,Fe3O4
Fe==>Fe(+3) + 3e
x 3x
Fe3(+8/3)==> 3Fe(+3) +1e
y y
N(+5) +3e ==> N(+2)
0.3 0.1
Ta có:
56x+232y=12
3x+y=0.3
==>x=0.09
y=0.03
vậy m=0.09*56+0.03*3*56=10.08
C2:
Fe ==>Fe(+3) +3e
x 3x
O2 +4e ==>2O(-2)
y 4y 2y
N(+5) + 3e ==> N(+2)
0.3 0.1
=> 4y +0.3=3x
56x+32y=12
=>x=0.18
y=0.06
==> mFe=0.18*56=10.08
C3:
nNO=0.1
nFe=m/56 => mFe(NO3)3=m/56
nHNO3= nNO3(-)+ nNO=3* m/56 + 0.1
nH2O= (3m/56 +0.1)/2
ĐLBTKL:
12 + [( 3m/56) + 0.1]*63=(242m/56) + 0.1*30 + {[(3m/56)+0.1]/2}*18
==> m=10.08
C4:
Quy đổi: Fe, FeO, Fe2O3, Fe3O4 thành Fe2O, Fe2O3
Fe2(+1) ==> 2Fe(+3) +4e
0.075 N(+2)
0.3 0.1
mFe2O= 0.075*128=9.6
mFe2O3=12-9.6=2.4
=> nFe2O3=0.015
==> m=( 0.075*2 + 0.015*2) *56=10.08
C5:
3FexOy + (12x-2y)HNO3 ==> 3xFe(NO3)3 + (3x-2y)NO + (6x-y)H2O
[3(56x+16y)]/12 = (3x-2y)/0.1
==> x/y=3/2 ==> Fe3O2
=> nFe3O2=12/200=0.06 => mFe= 0.06*3*56=10.08