High school chemistry trim, how to solve it? High one chemical trim

According to the charge conservation and balance, only the price of fe and n elements changes in the middle, fe from +2 to +3, n from +5 to +1, because the charge is conserved, so the product with the change of price level is allocated first, first assume that the balance is generated after N2O is 1, 2 n lose a total of 8 charges, Fe will get 8 charges, reactant feso4 will be preceded by 8, and because (SO4)2- all from feso4, it must be a multiple of 3, and the least common multiple of 3 and 8 is 24, Therefore, FeSO4 is preceded by 24, Fe2(SO4)3 is preceded by 8, and then according to the conservation of elements, Fe(NO3)3 is preceded by 8, at this step, Fe element will get 24 charges, and accordingly, there will be 6 n elements to get charges, and finally N2O is preceded by 3, and the rest can be balanced according to the conservation of the number of elements, 24FeSO4 + 30HNO3 = 8Fe(NO3)3 + 8Fe2(SO4)3 + 3N2O + 15H2O

24feso4+30hno3＝8fe(no3)3+8fe2(so4)3+3n2o+15h2o

The valence is conserved in rise and fall.

Fe2(SO4)3 increased Fe by 2 valence.

In N2O, N is reduced by a total of 8 valence.

Therefore, the coefficient of Fe2(SO4)3 can be 4

N2O has a coefficient of 1

After balancing, there are fractions, which are multiplied by 3

The 2 valence iron on the left becomes 3 3 valence iron on the right (moves 3 electrons) and the 5 valence n on the left becomes 2 1 valence n (moves 8 electrons), and the least common multiple of 24 shows that N2O is preceded by 3, and Fe(NO3)3 and Fe2(SO4)3 are preceded by 8

Seriously, I don't know whether the alkaline solution should produce water or iron hydroxide by oxygenation at the same time as ferrous and thiosulfate ions, and the first equation should be correct. Moreover, tetrasulfate is a very unstable thing, it is both oxidizing and reducing, in the case of oxygen introduction, can tetrasulfate exist stably? I reserve doubts.

However, this equation can actually be split into two parts, one is oxygen oxide ferrous ions, and the other is oxygen oxide thiosulfate ions.

The second equation is the +7 valent manganese ferrous oxide ion in the permanganate ion, +7 valence manganese becomes +2 valence, and +2 valence iron becomes +3 valence, which is relatively easy to balance.

Fe (+3) Fe (+6), liter 3 valence.

cl2(0) cl(-1), down 1*2=2 valence.

Follow Fe:Cl2

2:3 matching. In the order of fe and letter coarse cl na h stool silver o in turn with smooth and cautious elements:

2fe(oh)3

10naoh3cl2

2na2feo4

6nacl8h2o

After Fe3+ reaction, it is converted into +6 valence FeO42- and loses 3 electrons;

1 valent Cl is converted to Cl- after the reaction, 2 electrons are obtained.

The algebraic sum of electrons according to the gain and loss is 0

So Fe3 + front coefficient = 2, Clo - front coefficient = 3, so it is 2Fe(NO3)3+3KCLO+ KOH K2FEO4+ KNO3+ KCL+ H2O

According to the conservation of NO3- before and after the reaction, the front coefficient of Kno3 = 6 is obtainedAccording to the conservation of Cl element before and after the reaction, the front coefficient of Kcl = 3 is obtained, and according to the conservation of K element, the front coefficient of Koh = 2*2+6+3-3 = 10

Finally, according to the conservation of the H element, the coefficient in front of H2O = 5 So: 2Fe(NO3)3+3KCLO+10KOH==2K2FeO4+6KNO3+3KCl+5H2O

2Fe(NO3)3+3KCLO+10KOH 2K2FEO4+6KNO3+3KCl+5H2O, which seems to be flat. The method of electronic gain and loss has been forgotten ... Only feelings.