Ask for a problem about a chemical equation in junior high school

CO2 molecular weight: 44

H2O molecular weight: 18

Let xco2,yh2o be generated, then 44x:18y=11:9 so x:y=1:2

CH4 + 2O2 = CO2 + 2H2O to choose A

aco2:44 h2o:18

co2:h2o=11:9

The reaction yields 1 volume of CO2 and 2 volumes of H2O, so choose A

Let the number of carbon atoms be x and the number of hydrogen atoms be y, then 44x:9y=11:9, and the solution is x; y=1:4, so methane is right.

cxhy＋(x＋y/4)o2→xco2＋y/2h2o

cxhyoz+(x-y/4+z/2)o2→xco2+(y/2)h2o

Bring it in and know it, or list it one by one, the most rigid.

Let the moles of carbon dioxide and water be x and y, respectively

44x:18y=11:9

x:y=1:2 So the ratio of carbon dioxide to water is 1:2, and the ratio of carbon atoms to hydrogen atoms is 1:4a

Select a, calculate the ratio of the amount of matter from the mass ratio of carbon dioxide to water (carbon is 1 to 4 to hydrogen), so the compound contains one carbon atom and four hydrogen atoms in one molecule.

The mass ratio of carbon dioxide to water is 11:9, and if the ratio of the number of molecules to the two is x:y, then there is 44x:

18y=11:9 x: y=1:

2. From the chemical formula of carbon dioxide and water, it can be seen that the ratio of the number of carbon and hydrogen atoms after the reaction is 1:4, and the number of atoms before and after the reaction is unchanged, so the ratio of the number of carbon and hydrogen atoms in the compound is also 1:

4. Therefore, a methane is chosen.

Let the compound be cxhyoz

cxhy + 2x+y/2)/2 o2===x co2 + y/2 h2o

44x:9y=11:9

x:y=1:4

Then the compound is CH4 and choose A

It is rare to write a lot of chemical equations in the exam.

If you write it, it's also commonly used.

You don't necessarily get it if you don't use it often.

So just remember the usual ones.

You don't have to spend too much time trying to do anything else.

If you have time, you should deduce more experimental questions.

1 solution: (1) the mass of potassium iodate containing 20mg of iodine = 20mg So the answer is:

2) The quality of iodized salt that needs to be consumed every day ==

So the answer is:

Set the solute mass x in 10g of filtrate

bacl2+2agno3=ba(no3)2+2agcl208 287xx=

1) C%=2) Let the mass of barium chloride in the sample be y, then, 6+y

y = the mass fraction of barium chloride in the sample is:

Let the mass of the generated d-free be x and the mass of reactant b be y

a+ 3b=2c+ 2d

x, so x=, according to the law of conservation of mass silver balance, it is known that the mass of the reactant is equal to the mass of the combustion product.

So again because x=

So y = so the mass ratio of b to d, i.e. x:y is the front: 3

Let the relative molecular mass of C be 22, then the relative molecular mass of D is 9a + 3b = 2c + 2d

x y44/ y=

x=The mass ratio of b and c in the villain scum should be 11

The mass ratio of b and d in the reaction is :9

Problem solving process: The ratio of the relative molecular mass of C to D is 22 to 9, Min Huiqing generated gram C, and the stoichiometric number of C and D before the reaction bridge is equal, so it can be calculated to generate grams, so the total mass of the product is grams. According to the law of conservation of mass, the mass of b in the reactant should be grams, so b:

d is: 3.

First of all, according to the relationship between the chemical formula and the molecular formula of C and D, the law of conservation of mass can be obtained, and the reaction mass deficit of b is , then the mass ratio of b:d is ,::3

Solution: If 10 g of KCLO3 is completely reactive, let the mass of O2 it produce be non-reactive.

2kclo3 == 2kcl + 3o2

2* g 3*16*2 g

10g x g

2* g 10g = 3 * 16 * 2 g x g to obtain xg = , at this time it can be seen from the law of conservation of mass that the remaining solid mass should be: 12g, which is more than the kdok.

It may be due to the lack of kclo3, so there is a problem with the data of 10 grams of kclo3. The mass of KCLO3 should be greater than 10 grams.

Let the actual mass of kclo3 be yg. The remaining solid mass is grams, indicating that the mass of O2 produced is:

2kclo3 == 2kcl + 3o2

yg2* yg = 3 * 16 * 2, the solution is yg=, indicating that the mass of kclo3 should be greater than 10 grams.

2kclo3=2kcl+3o2↑

The amount of mass reduction and 02 is grams.

2* 2* 3*32x y can be found x=

y = so there is a problem with this data of 2 grams of mno2.

According to the topic, the mass of oxygen can be found: 10+, and the complete decomposition of 10 grams of kclo3 can only produce grams of oxygen, which is inconsistent, so when the complete reaction of the remaining solid should be 10+, if the gas generated is grams, then kcio3 is grams and the remaining solid is grams, so the remaining solids are problematic.

The remaining solid should be, less the mass of mno2! Generate O2 is.

You can give me an address, and I will send you all the equations in the third year of junior high school.

That's too much, mainly for oxidation reactions.

When the mass of the remaining substance decreases after the reaction, due to the formation of carbon dioxide gas, the mass of carbon dioxide is:

250+500-662==88g

Suppose the mass of calcium carbonate of the reaction is x

caco3+2hci=co2↑+cacl2+h2ox---88g

So x==200g

Mass fraction of calcium carbonate in limestone = = (200g 250g)*%80%.

Conservation of mass. CO2 quality is. 250+500-662

I think you can forget the rest.