19 MnO2 accounts for 25 in the mixture of 2gKClO3 and MnO2. After heating for a period of time, MnO2

Reaction equation: 2kclo3 --mno2) 2kcl + 3o2 There is a mixture of kclo3 and mno2 10g, where mno2 2g then kclo3 is 8g

According to the equation: (MNO2 is a catalyst and does not participate in the reaction, that is, the mass remains unchanged before and after the reaction) so after heating for a period of time and cooling, it is measured that the remaining solid MNO2 accounts for 25% because MMO2 has 2g, so the total solid mass = 2g 25% = 8g At this time, there is unreacted KCLO3 in the mixture, and the resulting KCl, the cause of the mass loss is the escape of the generated O2.

Let the mass of reacted KCl3 be Xg, and the mass of KCl generated by the reaction will be Yg2kClO3 --mNO2) 2kCl + 3O2 ......m--x---y---2g---2g

So the solution is: x = 245 48g =

y = 149/48g =

Before the reaction, there was a mixture of KCLO3 and MNO2 10g, of which 2g of mNO2 indicated that KCLO3 had a total of 8g

So the unreacted kclo3 mass after heating for a while = 8g - 245 48g = 139 48g =

Let the quality of O2 that can also be generated be ZG

2kclo3 --mno2）→ 2kcl + 3o2↑……m139/48g---z

Solution: z =

Here's the train of thought.

Solution: The mass of manganese dioxide is the mass of the remaining solid after the reaction is the mass of the solid reduction before and after the reaction.

then the resulting O2 mass is.

Let the mass of potassium chlorate participating in the reaction be x

2kclo3====2kcl+3o2↑

x245/96=x/

x=then the decomposition rate of potassium chlorate=

It's not all kcl, and kclo3 hasn't been broken down.

This is the quality of O2 produced.

2kclo3====2kcl+3o2↑xx=

The reduced mass is the mass that generates O2, yes.

The mass of potassium chlorate in the original mixture is x.

2kclo3=△、mno2=2kcl+3o2↑245 96

x245/x=96/

The mass of x=mno2 is .

The mass fraction of potassium chlorate is:

Answer: The mass fraction of KCLO3, MNO2 in the original mixture, and potassium chlorate in the original mixture is.

Reaction equation: 2kclo3 --mno2) 2kcl + 3o2 There is a mixture of kclo3 and mno2 10g, where mno2 2g then kclo3 is 8g

Calculated according to the equation: (MNO2 is a catalyst and does not participate in the reaction, that is, the mass remains unchanged before and after the reaction), so after heating for a period of time and cooling, it is measured that the remaining solid accounts for 25% of the MNO2, because the MMO2 has 2g, and the slag is based on the total solid mass = 2g 25% = 8g, at this time, there is unreacted KCLO3 in the mixture, and the resulting KCL, the loss of mass is due to the escape of the generated O2.

Let the mass of reacted KCl3 be Xg, and the mass of KCl generated by the reaction will be Yg2kClO3 --mNO2) 2kCl + 3O2 ......m--x---y---2g---2g

So the solution is: x = 245 48g =

y = 149/48g =

Before the reaction, there was a mixture of KCLO3 and MNO2 10g, of which 2g of mNO2 indicated that KCLO3 had a total of 8g

Therefore, after heating for a period of time, the mass of kclo3 = 8g - 245 48g = 139 48g =

Let the celebration also generate o2 of zg

2kclo3 --mno2）→ 2kcl + 3o2↑……m139/48g---z

Solution: z =

The brother judgment here only gives the idea, and the specific calculation is omitted.

There are 8g of potassium chlorate and 2g of manganese dioxide in the original mixture, and the oxygen production can be calculated when the potassium chlorate is completely decomposed.

After the reaction, manganese dioxide accounts for 25%, and the total weight can be calculated to be 8g, a decrease of 2g, and the reduction of 2g is the oxygen that has been released.

Using the complete pure decomposition of potassium chlorate, the oxygen production minus the amount of oxygen released can be calculated, and the amount of oxygen that has been released is the oxygen that can be produced.

The mixture of KCLO3 and MNO2 is 26 g. After the heating reaction is complete, the remaining solid mass is then 1What is the mass of O2 generation? 2.How many grams of mNO2 are in the original mixture?

Solution: The lightened mass of the solid matter is the mass of the oxygen produced.

2kClO3 = 2kCl + 3O2 (Conditions: Mno2 and heating) 245 96x

x = the mass of manganese dioxide is.

Mass fraction of potassium chlorate in the original mixture: [

1.The mass of oxygen produced is the mass of the solid reduction, yes.

2.Let the mass of kclo3 be x and the mass of kcl generated is y.

2kclo3=mno2、△=2kcl+3o2↑245 149 96

x yx = y= kclo3 in the mixture.

3.The mass of mnO2 in the original mixture is:

From the calculation of 2 questions, it is known that the mass of KCL is.

The remaining solid contains Mno2 and KCl, where Mno2 is and KCL is.

(1) The reaction of KCLO3 and Mno2 produces only solids and oxygen, oxygen =

2) 2kclo3==2kcl+3o2 oxygen has, reverse calculation of potassium chlorate, according to the relative molecular weight.

3) In the second step, after calculating the mass of potassium chlorate, subtract it to the mass of mno2, and subtract the mass of mno2 to the mass of potassium chloride. What remains is a mixture of potassium chloride and mnO2.

The mass of the solid decreases, and the mass of the reduction is the mass of O2.

Mass of O2 = - = g

Let the mass of KCL be x

2 kclo3 ==△mno2== 2 kcl + 3 o2↗· 149 96

x g149/96 = x / g

So x = g

That is, the mass of the KCl generated is g

2kclo3 = 2kcl + 3o2

x = 149 * y = A grams of oxygen generated in grams of potassium chloride.

Let the mass of the original mixture be 1 and the mass of oxygen generated is x, and the mass of the remaining solid plate can be obtained as 1-x, which is unchanged before and after the catalyst mass reaction, so: 1*20%=(1-x)25% solution x=, substituted into the equation.

2kclo3=2kcl+3o2

y solves y=, the decomposition rate of kclo3=

2kclo3===heating destroys the first letter ==mno2==2kcl+3o2

KCLO3 reacts completely, so only the product KCL and the catalyst Mno2 are left in the solid, because the mass of the catalyst remains unchanged before and after the chemical reaction, so the mass of the fiber wheel Mno2 is still 2 grams.

So, m(kcl) = gram celery.