A chemical equation bulldozing problem

feso4 + hno3 － fe(no3)2 + fe2(so4)3 + n2o + h2o

First of all, this is a redox reaction.

It can be leveled by the valence lifting method:

1.R (reducing agent) FeSO4 -> Fe2(SO4)3 +1

O (oxidant) Hno3 -> N2O -4

Cross to get 4 FeSo4 + 1 Hno3

2.According to Fe, N material conservation pre-matching:

4 feso4 + 1 hno3 --2 fe2(so4)3 + 1/2 n2o

Multiply by 2 to eliminate the fraction:

8 feso4 + 2 hno3 --4 fe2(so4)3 + 1 n2o

3.At this time, a large anion (SO4)2- is added to conserve the material

8+4)feso4 + 2 hno3 --4 fe2(so4)3 + 1 n2o

Then conserve the Fe2+ charge (with Hno3).

12 feso4 + 2+4*2)hno3 --4 fe(no3)2 + 4 fe2(so4)3 + 1 n2o

4.Finally, H2O is used to balance H and O

12 feso4 + 10 hno3 --4 fe(no3)2 + 4 fe2(so4)3 + 1 n2o + 5 h2o

5.Use the material conservation check, omit the coefficient "1", write "=" to indicate that it has been balanced, mark the gas, precipitation, write the reaction conditions, etc.

12 FeSO4 + 10 Hno3 == 4 Fe(NO3)2 + 4 Fe2(SO4)3 + N2O(gas) +5 H2O

The trim equation must first be patient and careful, then be sober-minded, and finally be sure to check, in fact, the trim equation is the essence of chemistry, there can be difficult problems, but practice makes perfect, more practice will have a feel.

All right. Now that they're all in. Tell me about my thoughts.

First of all, observe the valence change There is nitric acid in this formula, n is reduced, from.

5 to +1 (n20).

Note that some of the ferric iron has no valence change, and has become ferrous nitrate, leaving him alone for the time being.

Some of the iron changed from 2 valence to 3 valence, and the 1 valence increased, so there is a 4:1 relationship (conservation of electrons) between the product and the product of n just now

Well, this is the end of the redox analysis, and it begins to accompany.

According to 4:1, it is observed that although the iron and nitrogen in the product are two together, the ratio is the same, so write 4 in front of the sulfate of ferric iron and 1 in front of nitrous oxide

After that, find elements that appear less often to facilitate trimming.

It is observed that there are 3*4=12 sulfate in the product, and according to the conservation rule, write 12 at the ferrous sulfate in front

This makes it possible to further write the coefficient of ferrous nitrate of the product.

is 12-2*4=4, write 4 before ferrous nitrate

Then look at the number of nitrogen elements: ferrous nitrate contains 2 nitrogen, and 1 oxide 2 nitrogen also contains 2.

The coefficient is known, so the total amount of nitrogen is 4*2+1*2=10 before nitric acid is written 10 and finally 5 water is written with the conservation of hydrogen. Verify.

Answer. To say how to calculate.

Redox is basic.

fe +2 to +3 increased by 1 x 4

n +5 to +1 decreased by 4

fe there is to be multiplied by 4 valence to flat).

And then this is 4 is added at the ferric sulfate.

And the coefficient of n2o is 1, don't move.

After adding this 4, you can use it later.

Observation is balanced.

One last word.

I'm not very good at chemistry.

Maybe it's a stupid approach.

To find a way, it's best to look at reference books.

There are very detailed steps and some necessary rules.

Of course, you have to practice more yourself.

It is redox that reflects the gain and loss of electron equilibrium.

Is it enough to simply make up the direct conservation of mass? The complex redox reaction equation is supplemented by the conservation of electrons and charge gains.

For example, MnO4- +C2O4(2-) potassium permanganate oxidizes oxalate in an acidic environment.

First of all, you need to know what a redox product is. When acidic, the reduction product of permanganate is divalent manganese, and the oxalate oxalate oxalate oxalate oxalate is carbon dioxide. Then use the conservation of electrons of gain and loss, manganese in permanganate 7 valence, to divalent manganese down five valence, oxalate carbon positive 3 valence, carbon dioxide positive 4 valence, liter one valence, one oxalate two carbon, so one oxalate root rises bivalent.

So permanganate:oxalate is 2:5, and that's right.

2mno4- ＋5c2o4（2-）＝2mn2+ +10co2

Then with the conservation of charge, the left side is totaling -12, the right side is totaling +4, so in an acidic environment, 16 hydrogen is added on the left and 8 water is added on the right. That is.

2mno4- +5c2o4（2-）+16h+ ==2mn2+ +10co2+8h2o

I still have to practice, and I feel that all equations are the same.