In the 2KMnO4 16HCl 2KCl 2MnCl2 5Cl2 8H2O reaction the reducing agent is , and the reducing product

Cu+4Hno3=Cu(NO3)2+2NO2+2H2O, the oxidant is the HNO3 oxidation product NO2, the reducing agent is Cu, and the reduction product is Cu(NO3)2

Cu+4Hno3=Cu(NO3)2+2NO2+2H2O oxidant is HNO3 The oxidation product is Cu(NO3)2, the reducing agent is Cu, and the reducing product is NO2

2kmNO4+16HCl=2KCl+2MnCl2+5Cl2+8H2O The oxidant is KMno4, and the oxidation product is Cl2

Note: The oxidant is reduced, corresponding to the reduction product.

The reducing agent is oxidized, corresponding to the oxidation products.

What is produced by the oxidation of the reducing agent is not an oxidation product.

What is produced by the reduction of the oxidant is not a reduction product.

The first reaction formula, oxidant HNO3, oxidation product Cu(NO3)2, reducing agent Cu, reducing product 2NO2, the second reaction formula, oxidant KMno4, oxidation product MnCl2, reducing agent HCl, reducing product Cl2, in the reaction, the oxidant is reduced, generating a reduction product, the valence drops, the reducing agent is oxidized, the oxidation product is generated, and the valence rises.

The oxidant agent is 4Hno3, the oxidation products 2NO2 and 2H2O, the reducing agent Cu, and the reducing product Cu(NO3)2

According to the reaction equation 2kmNO4+16HCcl(concentrate)->2KC+2MnCl2+8H2O+5Cl2, 2 KMnO4 were returned to the original missing orange spike with 2 mnCl2, so the amount of oxidized volvo was 2mol.

In the 1mol reaction formula, KMno4 is the oxidant, 16HCl (concentrated) is the reducing agent, chlorine gas is prepared to obtain chlorine gas, that is, chlorine gas, and a total of 5mol electrons are obtained, so the oxidant needs to lose 5mol of electric fibers, and 1mol KMno4 becomes 2mnCl after the reaction, which is just like the loss of 5mol electrons The answer is 1mol

The valency of KMno4 is reduced by Mn, so the increase in the valency of KMno4 as an oxidant is Cl, so HCl is used as a reducing agent.

16 HCl participated in the reaction, and only 10 had a change in valency. In other words, there are 10 HCl that are oxidized.

2kmNO4+16HCl (concentrated)=2KCl+2MnCl2+8H2O+5Cl2

x 71/71=1

x = 2 mass of oxidized HCl = 2*

The oxidant is KMNO4

The reducing agent is HCl

First, calculate the n=m m=71 71=1mol of 71 grams of chlorine and set the pseudoxmol of hydrochloric acid consumed

According to the equation, the coefficient ratio is 16HCl 5Cl2 hydrochloric acid, but the oxidized is 10molCl ions: chlorine is 10:5 and can be listed according to the ratio of 5 1=10 X

Calculate x=2mol m=m(hydrochloric acid)*n(hydrochloric acid)=(

The oxidized element is Cl, there are a total of 16 Cl, but the true valency change, it becomes chlorine Cl2 is indeed only 10, that is, there are only 10 Cl participating in redox, and Mn has two valency changes, all participate in redox, so it is 10:2=5:1

16 HCl molecules participated in the reaction to generate 5 Cl2 molecules, that is, 10 Cl- ions were oxidized into atoms, and the other 6 Cl- participated in the reaction as acid groups. There are 2 mnO4- ions that are reduced to 2 Mn2+. So:

The ratio of the amount of matter to the element being oxidized is 10:2=5:1

7 -1 -1 +2 -1 0 2kmno4+16hcl=2kcl +2mncl2+5cl2+8h2o

Both mn7+ prices have been changed to 10E-

6 of the 16 cl- did not change the price and lost 10 e-

So NHCL (variable): nkmno4 (variable) = 10:2 = 5:1

Because not all Cl is oxidized, the chlorine in manganese chloride and potassium chloride is not oxidized.

Can't count.

Observing the chemical equation, Mn changes from +7 valence to +1 valence, and KMno4 is the oxidant. Cl changed from -1 valence to 0 valence, and HCL was the reducing agent; The ratio of the two is the ratio of stoichiometric numbers, which is 1:8. The reducing agent is oxidized and has oxidation products, which are Cl2; The oxidant is reduced, and there is a reduction product, which is MnCl; The ratio of the two is the ratio of stoichiometric numbers, which is 5:2

The ratio of oxidation products to reduction products cl2: mncl2 = 5:2

The ratio of oxidant to reducing agent kmno4:HCl=2:10

Oxidation products are more than reduction products = 5|1

Oxidant ratio to reducing agent = 1|5

The oxidant is potassium permanganate, and the valency of manganese is all reduced, so the oxidant is 2mol; The reducing agent is HCL, look at the right side of the equation, only 10mol of chlorine has an increased valence, so the reducing agent is 10mol HCl. Therefore, the amount of the reducing agent is 10mol, and the ratio of the amount of oxidant to the reducing agent is 2 to 10, which is 1 to 5

The oxidant is potassium permanganate, which is 2mol; The reducing agent is HCl, but only 10mol of 16mol has an increased valence, which is a real reducing agent. Therefore, the amount of the reducing agent is 10mol, and the ratio of the amount of oxidant to the reducing agent is 2 to 10, which is 1 to 5

KMno4: Mn valence is +7 MnCl2 valence is +2 So potassium permanganate is an oxidizing agent.

Cl is -1 and Cl2 is 0 in hydrochloric acid, so hydrochloric acid is the reducing agent, and there are 10 moles of chlorine that undergo valence change, because 5 moles of chlorine are generated.

In summary, it is 2:1

Answer: (Chlorine) element is oxidized - the valency of chlorine is from 1 liter to 0 valence (KMNO4) is an oxidant - the valency of manganese is reduced from +7 to +2 (Cl2) is an oxidation product - HCL is oxidized and converted to Cl2 (HCl) Oxidation reaction occurs - the valency of chlorine is increased by 2 mn from +7 to +2, 10 electrons are transferred, and 5 Cl2 (1mol Cl2 corresponds to 2mol electrons).

So for every 1molCl2 produced, 1 5*5=1mol electrons are transferred.

1. The oxidant is 2kmNO4

The reducing agent is HCl (concentrated).

The HCl involved in the redox reaction is 10HCl

The ratio of the amount of the oxidizing agent to the reducing agent is 1:5

2. Concentrated HCl reflects acidity and reduction.

3. The molar mass of KMnO4 is 158g mol, so N(KMno4)=From the equation, KMno4 is better than HCl=2 to 16, so the HCl participating in the reaction is M(HCl) = if it is oxidized, HCl=

4. The ratio of the amount of oxidant to the reducing agent is 1:5, and the ratio (chlorine) = 2 to 1 (HCl) is oxidized.

5. When KMNO4 is 2mol, the transfer electron is 10mol. So when there is electron transfer in the reaction, chlorine gas is formed.

v=6, generated, then KMNO4 participates in the reaction so there are 2mol electrons transferred.

1. The oxidant is.

2kmno4

The reducing agent is HCl (concentrated).

HCl involved in the redox reaction is.

The ratio of the amount of 10HCl oxidant to reducing agent is 1:52, and the concentrated HCl reflects acidity and reduction.

3. The mass of KMNO4 is 158g mol, so N(KMNO4)=

It is known from the equation.

KMNO4 vs. HCL=2 vs. 16

So the slag HCl that participates in the reaction is.

The oxidized HCl is.

m (oxidized HCl) =

4. The ratio of the amount of oxidant to reducing agent is 1:5

HCl) ratio (chlorine) = 2 to 1, so when oxidized HCl, chlorine can be generated.

5. When KMNO4 is 2mol, the transfer electron is 10mol. So when there is electron transfer in the reaction, chlorine gas is formed.

v=6, generated, there is.

KMNO4 participates in the reaction.

So there are 2mol of electrons that are transferred.