Coi X gồm m gam Fe và (3-m) gam O
=> nFe= x/56 mol; nO= (3-m)/16 mol
nNO= 0,56/22,4= 0,025 mol
Fe -> Fe+3 +3e
=> n e nhường= 3m/56 mol
O+ 2e -> O-2
N+5 +3e -> N+2
Ta có pt: 0,025.3+ 2(3-m)/16= 3m/56
<=> 0,075+ (3-m)/8= 3m/56
<=> (0,6+3-m)/8= 3m/6
<=> 24m= 6(3,6-m)
<=> m= 0,72g