C2H5OH KMnO4 H2SO4 CO2 K2S04 MnSO4 H2O How to trim

In organic matter, the valency change is generally element C. Element c is taken as the average valency.

The valency of C in C2H5OH is -2 valence. The valency of C in CO2 is +4A C increases by 6 valence, for a total of 12 valence.

MN in KM4 is reduced by 5.

So the raised ones are multiplied by 5 and the lower ones are multiplied by 12

5C2H5OH+12kmNO4+H2SO4=10CO2+K2S04+12MnSO4+H2O, and then the observation method can be balanced.

5c2h5oh+12kmno4+18h2so4=10co2+6k2s04+12mnso4+33h2o

MN changes from positive 7 to positive 2, decreasing by 5; c changes from -2 to +4, up 6

5 c2h5oh+12 kmno4+18 h2so4=10 co2+6 k2s04+12 mnso4+33 h2o

The key to the balancing of the chemical reaction equation is the gain and loss of electrons, which is manifested in the change of the price state of the left and right oxides being reduced. Grasp the core change, and the rest of the work is just to make the total number of atoms on both sides unchanged.

Looking at both sides, it can be seen that the c is oxidized in the formula, and the valence state is -2 to +4, and 6 electrons are lost, but for the sake of trim convenience, ethanol has 2 c, so it loses 10 electrons for an ethanol molecule; Mn is reduced, from +7 to +2, 5 electrons, 5 and 12 least common multiple 60, so fill in 5 before the left ethanol, fill in 12 before potassium permanganate, and the rest of the work only needs to ensure that the number of atoms on the left and right is equal.

The answer is 5, 12, 18, 10, 6, 12, 33

The trim is 2kmNO4+3 H2SO4+5H2C2O4=10CO2+ 2 MnSO4+ K2SO4+8H2O.

KmNO4——— MnSO4, the valency of Mn changes from 7 to 2, and 5E-2 is transferred.

H2C204-2CO2C2, the valency changes from two 3 to two, and 2e- x5 is transferred.

According to the conservation of electrons of gain and loss, the least common multiple of the number of electrons transferred is 10, so the number of measurements is matched separately, and the matching is as follows:

2kmNO4+ H2SO4+ 5H2C2O4=10CO2+ 2 MnSO4+ K2SO4+H2O, and then according to the number of potassium ions, K2SO4 is preceded by 1, then H2SO4 is preceded by 3, and H2O is preceded by 8, and 2kmNO4 + 3 H2SO4 + 5H2C2O4=10CO2+ 2 MnsO4+ K2SO4 + 8H2O.

Basic principles: 1. The number of atoms of each element before and after the reaction is equal, that is, the mass is conserved.

2. The total number of electrons obtained by the oxidant is equal to the total number of electrons lost by the reducing agent, that is, the electrons are conserved.

3. The total value of the decrease in the valency of the oxidant is equal to the total value of the increase in the valency of the reducing agent.

Please adopt it promptly if you understand!! Good luck with your studies!!

The ratio of the number of measurements: 2:3:1:2:2:1:4

It is possible to use the valence rise and fall method or write it as a semi-reaction.

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

First, let the front coefficient of C2H5OH be 1, the front coefficient of KMNO4 be rotten X, and the front coefficient of H2SO4 be Y, and then make a two-liquid calendar and balance of equations about X and Y, and solve X and Y

8 h2o + 5 k2s2o8 + 2 mnso4 ==4 k2so4 + 2 kmno4 + 8 h2so4

According to the valency change.

Actually S2O8- is +6 valence in S because sulfur has only 6 valence electrons. Two of the oxygen are -1 valence.

1. The first standard is the only way to get out of the price of the synthetic ant

mnso4+ k2s2o8+ h2o=kmno4+ k2so4+ h2so4

Rise 5 drop 22, with the least common multiple of the ligand to level the redox part:

2 mnso4+ 5 k2s2o8 + h2o ==2 kmno4+ 4 k2so4 + h2so4

3. Balance other atoms by observation:

2 mnso4+ 5 k2s2o8 + 8 h2o ==2 kmno4+ 4 k2so4 + 8 h2so4

The sulfur valency of SO2 rises from +4 to +6, rises by 2, and the valency of manganese with 5kmNO4 decreases from +7 to +2, decreases by 5, and then according to the conservation of potassium and manganese, with potassium sulfate is 1, manganese sulfate is 2 and then according to the conservation of sulfur, with sulfuric acid is 2

Then according to the conservation of the hydrogen element, the water distribution is 2

Check whether the oxygen element is conserved before and after, 20 on the left and 20 on the right, conserved.

So the end result is:

5so2+2kmno4+2h2o=k2so4+2mnso4+2h2so4

Mn is +7 valence in KMno4 and +2 in MNSO4, giving 5 electrons.

The S-2 valence in K2S, the S+6 valence in SO42- in the product, and the number of uncleared electrons is 8 electrons.

Then the kmNO4 coefficient is 8, the K2S coefficient is 5MnsO4, the K2SO4 coefficient is 8, the K2SO4 coefficient is 5H2SO4 coefficient, and the positive base is 8+5==13

H2O is 13

i.e. 8kmNO4 + 5K2S +13H2SO4 = 8mNO4 + 5K2SO4 + 13H2O

2kmno4+5k2c2o4+8h2so4=10co2+2mnso4+8h2o+6k2so4

You can first match the ion equation Wu Qinghui 2mNO4- +5C2O4- +16H+ =2MN2+ +10CO2 + 8H2O, and then add K+ and differential SO42-

The oxidant KMNO4 was reduced to MnSO4, and manganese was reduced by 5.

The reducing agent H2S is oxidized to S, which increases the 2-valent.

So KMNO4 is preceded by 2, and H2S is preceded by 5

Observe the trim of other substances.

2kmno4+5h2s+3h2so4==k2so4+2mnso4+5s↓+8h2o

In junior high school, then use the observation method:

Look at potassium first, two on the right, one on the left, then KMNO4 with 2, then look at manganese, two on the left, and another one, then MNSO4 with 2, and then look at sulfur, three on the left, then H2SO4 with 3, and then look at hydrogen, with 3 sulfuric acid, then the water behind has 3 from sulfuric acid, there are n from hydrogen peroxide, with n H2O2, (N+3) H2O, and then look at oxygen, the oxygen behind should come to the dust banquet from the hydrogen oxygen jujube silver water, then there are n hydrogen peroxide, the left and right oxygen atoms should be equal, then Yanzhen has 2*4+n*2+3* 4=4+2*4+(n+3)*1+n*5, the solution is n=5, and then ok