Can anyone help me find 30 balanced chemical formulas for emergency

feo42- +h2o ——fe(oh)3 + oh- +o2

FeO42- is called high-iron acid heel, where Fe is +6 valence, and Fe is +3 valence in Fe(OH)3 precipitation, and each Fe atom gets 3E-;

O in H2O is -2 valence, and O in O2 is 0 valence, and each O atom loses 2 electrons to form an O2 molecule with 4E-.

According to the total number of electrons gained and lost, the least common multiple is 12, then FeO42- and Fe(OH)3 are multiplied by 4, and O2 is multiplied by 3

4feo42- +h2o ——4fe(oh)3 + oh- +3o2

The reason why the coefficient is not written for H2O is because the O atoms do not all become 0 valence, but only the O atoms in the generated O2 that become 0 valence).

Then according to the conservation of charge, 4 FeO42- on the left with 8 negative charges, H2O is not charged, and there must also be 8 negative charges on the right, then oh- times 8

4feo42- +h2o ——4fe(oh)3 + 8oh- +3o2

Then it can be leveled according to the conservation of mass.

4feo42- +10h2o == 4fe(oh)3↓ +8oh- +3o2↑

Iron in ferratic acid is +6 valence, iron hydroxide is +3 valence, oxygen in water molecules is -2 valence, and oxygen in oxygen is 0 valence, so the following chemical equation skeleton can be written:

4 ferroic acid (2-)+6 water molecules + ......=4 Iron hydroxide + 3 oxygen + ......

Now the left side of the equation is 22 oxygen, the right side is 18 oxygen, the electrons are 8 on the left, and the right side is 0, because this is an alkaline environment, so you can only take water and hydroxide to balance, so add 8 hydroxide on the right side of the equation, and then balance the atoms with water, so add 4 water molecules on the left, so you get the equation shown in the diagram.

+7 valent Mn in potassium permanganate becomes +2 valence Mn in manganese sulfate to obtain 5E-. +2 valent Fe in ferrous sulfate becomes +3 valent Fe in ferric sulfate, losing E-. Then there must be 2 mns to gain 10e-, and 10 fe to lose 10 electrons, so that the number of electrons gained and lost can be conserved.

Then 2kmNO4+10FESO4+8H2SO4=5FE2(SO4)3+2MNSO4+K2SO4+8H2O

Mn+7 in KMNO4 and Mn+2 in MNSO4 decreased by 5

Fe+2 in FeSO4 Fe2(SO4)3 increased by 1

2kmno4+10feso4+8h2so4=5fe2（so4）3+2mnso4+k2so4+8h2o

Let's take a look at the law of conservation of mass, do a few more questions, ask teachers, classmates, it's easy, let parents talk about it.

It's really not possible to buy a reference book to see, no matter how much others say, it's useless, and you really understand it yourself!

Hehe, I wish you a speedy figuration of this!

Let = have an equal number of atoms on both sides, H2+O2 - H2O is 2H2+O2====2H2O after balbining, and there are 4 hydrogen atoms and 2 oxygen atoms on both sides.

By the time you're in high school, it's easy. And then I did more, and those methods of chemical balancing in junior high school can really be said to be case experience.

...You can ask your teachers again.

Your equation is a typical redox reaction, and the specific balancing method is as follows:

mnso4+k2s2o8+h2o--h2so4+k2so4+kmno4

First, observe which element in the equation has a change in valence, in this case, Mn rises from +2 valence to +7 valence, transferring 5 electrons;

The valency of the element decreases, and then looks for which element has the valency decreased, one look at the valency of the S element from +7 to +6, transferring 1 electron, to make the number of electron transfers equal, it must be multiplied by 5 (similar to the cross in mathematics), so the coefficient in front of K2S2O8 is 5, and the coefficient in front of MnSO4 is 2 (note that 2 here instead of 1 is because there are two Ss in K2S2O8, and each S transfers an electron for a total of two). I think you'll be able to level the rest of the rest with ease. Good luck!

MN: up 5 prices. *2

2 o's each drop 1 price. A total of 2 prices were reduced. *5

5c2h5oh+12kmno4+18h2so4==2co2+6k2so4+12mnso4+33h2o

Procedure: First, let the front coefficient of C2H5OH be 1, the front coefficient of KMno4 be X, and the front coefficient of H2SO4 be Y, and then find the binary linear equations about X and Y, and solve X and Y

Step 1: Suppose the coefficient of Na2O2 is 1; Then the coefficient of NaOH is 2Na2O2+H2O --2NaOH+O2

Step 2: The coefficient of water is 1, count the O atoms, then O2 with 1 2 Step 3: Remove the denominator, get: 2Na2O2+2H2O = 4NaOH+O2 Electron transfer:

The oxygen in sodium peroxide rises from -1 valence to 0 valence and becomes an oxygen atom (i.e., an oxygen molecule), and the other decreases from -1 to -2, so the sodium peroxide gives itself electrons. So the stoichiometric number is 1

According to the conservation of sodium atoms, sodium hydroxide is preceded by two to obtain:

na2o2+h2o=2naoh+（

Sort it out and you've got it.

2na2o2+2h2o=4naoh+o2↑

2Na2O2+2H2O =4NaOH+O2 Step 1: Assuming that the coefficient of Na2O2 is 1; Then the coefficient of NaOH is 2 Step 2: the coefficient of water is 1, count the O atoms, then O2 with 1 2Na2O2 + H2O = 2NaOH + 1 2O2 Both sides of the equation are multiplied by 2 to obtain:

2na2o2+2h2o = 4naoh+o2↑

First with na, then with o, and finally with h

It comes out with 1,1,2,1 thank you.

In fact, the trimming of chemical equations is not to find any common factor, but it is easy to match after you really know its electron transfer, there is no process, high school chemistry actually did not learn anything, just learn a conclusion.