Trim the following chemical equation 10

The chemical equation of trimming first looks at the type and number of atoms on the left and right sides.

1) 1 w 3 o 2 h on the left

1 W 2 H 1 O on the right

First, add 3 before H2O on the right to make the number of oxygen atoms the same.

Left 1 w 3 o 2 h

On the right, 1 w, 3 o, 6 h

Add 3 before h2 to make the number of hydrogen atoms the same.

Left 1 w 3 o 6 h

On the right, 1 w, 3 o, 6 h

So it's 1:3:1:3 (1)wo3+(3)h2 heats 1)w+(3)h2o

2) Left 1 C 4 H 2 O

On the right, 1 c, 3 o, 2 h

Trim next to odd numbers by adding 3 before CO2, then 3 before CH4, and finally 6 before H2O, and 6 before O2

3:6:6:3=1:2:2:1 (1)CH4+ (2)O2 ignites 2)H2O+(1)CO2

2kmNO4==Heating==K2mNO4+MNO2+O2

1) (1) WO3+(3)H2 heating 1)W+(3)H2O2) (1)CH4+ (2)O2 ignition 2)H2O+(1)CO2

K2mno4==Heating==KMno4+MNO2+O2 is changed to 2kmNO4==Heating==K2mNO4+MNO2+O2 According to the valency, MN4 is +7 in KMno4, MNo2 is +4, and MN2 is +6 in K2mNO4

O in KMno4 is -2 There is 2 in O2 in O (is 0 valence) 0 From -4 to 0 there are 4 electrons transferred in KMno4 Mn has a change of +4 and one of them becomes +6

There should also be 4 electron transfers.

1) (1) WO3+(3)H2 heating 1)W+(3)H2O

2)( CH4+ (2)O2 ignition 2)H2O+( CO2 corrects the following chemical equation.

2kmNO4==Heating==K2mNO4+MNO2+O2

Coefficients: 1,3,1,3

The third one you obviously wrote wrongly, you wrote potassium permanganate and potassium manganate in reverse, and the coefficient is: 2,1,1,1

1、 yco+ fexoy== xfe+ yco2

2、 yc+ 2fexoy== 2xfe+ yco2

3、 yh2+ fexoy== xfe+ yh2o

4. 4cxhy+ (4x+y)o2==4xco2+ 2yh2o These are the general methods and steps of redox reaction, redox reaction equation balancing.

1. General method: from left to right.

2. Steps: mark the price, find the change, find the total number, and match the coefficient. Namely.

The beginning and end states of the valency of the changing elements are marked;

The total number of values that change in the beginning state and the final state = coefficient of change.

Note: It is assumed that the above changes are expressed in positive valence, where (b-a) (d-c) is the least common multiple.

The coefficient on is filled in in front of the chemical formula of reducing agent and oxidant respectively as coefficient;

Trim other elements by observation;

Check whether the equation after trim conforms to the law of conservation of mass (the ion equation also depends on whether it conforms to the conservation of charge).

Example 1: C + Hno3 (concentrated) - NO2 + CO2 + H2O

Analysis: Bid Variation.

C(0) +hn(+5)O3(concentrated)- N(+4)O2 + C(+4)O2 + H2O

Look for changes. The total number of values that change in the beginning state and the final state = coefficient of change. Find the total.

Matching coefficient. The coefficient of c is 1 and the coefficient of hno3 is 4, and the other coefficients are balanced by observation.

Finally, it is checked to meet the law of conservation of mass. The chemical equation after trimming is:

C + 4 Hno3 (concentrated) = 4 No2 + CO2 + 2 H2O

1, y 1 x y 2, y 2 2x y 3, y 1 x y 4, there is a formula that I learned in the organic chapter, I forgot.

(1) 2NaOH + CuSO4 - Cu(OH)2 down arrow + Na2SO4

2) Fe2O3+3H2—High temperature—2Fe+3H2O

3) Fe2O3+3Co—High temperature—2Fe+3CO2

(1) 2 + 1 (1 to be omitted) = 1 (1 to be omitted) + 1 (1 to be omitted).

2) 1 (1 to be omitted) + 3 = 2 + 3

3) 1 (1 to be omitted) + 3 = 2 + 3

precedent with 2; 2。1, 3, 2, 3;3。1, 3, 2, 3;

1.()CH4+(2)O2---2)CO2+(2)H2O(conditional ignition)2（）so2+(2h2s)--3)s(↓)h2o3.

2) NaOH+CUSO4---NA2SO4+CU(OH)2 rebates.

4.(laughing world chaos 2) touches al+(3)h2so4---al2(so4)3+(3)h2

5.（2）fe+（2）hcl---fecl2+h2↑

Redox reaction balancing law: according to the change of valency, take the least common multiple of the increased and decreased valence to match, and you can do it.

s+o2==so2

ch4+2o2==co2+2h2o

8al+3fe3o4==9fe+4al2o32h2s+so2==2h2o+3s

mno2+4hcl==mncl2+2h2o+cl22c2h6+7o2==4co2+6h2o

1.It has been trimmed.

fe4o3 is not a mistake).

(1) C2H4+O2—CO2+H2O (condition: ignition).

Starting from C2H4, first balance the c atoms on the left and right sides of the equation with the h atoms to get:

c2h4+（）o2—2co2+2h2o ；

Then according to the number of oxygen atoms on the right is equal to the number of oxygen atoms on the right, fill in 3 in parentheses to get:

c2h4+3o2—2co2+2h2o

2）naoh+cuso4—cu(oh)2↓+na2so4

The number of Cu ions and SO4 ions on the left and right sides of the equation is already equal, and the Na ions and OH ions are mainly considered. There are 2 Na ions and OH ions on the right side of the equation, so you only need to put 2 in front of NaOH on the left side of the equation:

2naoh+cuso4—cu(oh)2↓+na2so4

3) Fe2O3+H2—Fe+H2O (condition: high temperature).

This is a redox reaction that should be considered from the rise and fall of valency. o element is -2 valence on the left and right sides of the equation, so let's not consider it for now. The Fe element changes from 3 valence to 0 valence, and the H element changes from 0 valence to 1 valence, indicating that 1 part of Fe needs 3 parts of H to reduce, so 2 parts of Fe need 6 parts of H to reduce, so to reduce two parts of Fe in a Fe2O3 molecule, 6 H atoms are needed, that is, 3 H2 molecules.

So in front of h2 with 3 gets:

fe2o3+3h2—（）fe+（）h2o。

Then according to the relationship that the number of FE on the left and right sides of the equation is equal, 2 is assigned in the front brackets of FE on the right, and 3 is matched in front of H2O according to the relationship between hydrogen or oxygen left and right, and 3 is given in front of H2O to obtain:

fe2o3+3h2—2fe+3h2o。

4) Fe2O3+CO—Fe+CO2 (condition: high temperature).

FE element changes from 3 valence to 0 valence, C element changes from 2 valence to 4 valence, and one FE element needs three-thirds of C element to reduce, that is, 2 FE needs 3 parts C, and one FE2O3 needs 3 CO. So on the left co, preceded by 3:

fe2o3+3co—（）fe+（）co2

Considering the number of Fe and C atoms on the left and right sides of the equation, 2 is assigned before Fe on the right and 3 before CO2

fe2o3+3co—2fe+3co2

(1) 4al + 3o2 ＝ 2al2o3

2) 1Fe3O4 + 4Co High Temperature 3Fe + 4CO2

3) 1fe2o3+ 6hcl ＝ 2fecl3 + 3h2o

4) 1Al2(SO4)3 + 6NaOH = 3Na2SO4 + 2Al(OH)3 (precipitation symbol).

5) 4FeS2 + 11O2 High temperature 2Fe2O3 + 8SO2

This is a special chemical equation trimming problem, so when the stoichiometric number is "1", it cannot be saved.

Your 10 points of satisfaction, the infinite motivation of our team].

1) 2 C2H2+ 3 O2 ignition to generate 2 CO2+2 H2O2) Fe2O3+2 CO high temperature to generate 2 CO2+ 2 Fe3) 2 Al+ 3 H2SO4 to Al2(SO4)3+ 3 H2

4) 4 NH3+ 5 O2 High Temperature Generation 4 NO+ 6 H2O Note: Equation (3) is incorrect and has been modified.

c2h2+

o2 ignites with vertical noise.

co2+2h2o

fe2o3+2

CO high temperature generation.

co2+feal+

H2SO4 generation.

Al2 (dust SO4) 3+

h24) Talk.

NH3+O2 is generated at high temperature.

No+H2O Note: Equation (3) is incorrect and has been modified.