High school chemistry calculation problems. From the textbook of the second science class of Hua II

First of all, you have the wrong ...... in your first question

Get to the point: There are two solutions to the first question: 1) CO excess: According to 2CO+O2=2CO2, it can be seen that the reduced 1ml is the volume of O2.

After passing through the NaOH solution, the volume is reduced by 5ml, indicating that the volume of CO2 after ** is 5ml, and the starting CO2 is 3ml from the generation of 2ml, and the remaining gas is CO, plus the first 2ml, it can be known that the starting CO is 5ml

2) O2 overdose The solution is similar, from 2CO+O2=2CO2 to know that CO is 2ml, CO2 is unchanged to 3ml, and O2 is the remaining 3ml, plus a small consumption, one ml is 4ml

Question 2 MNO2 is useless for catalyst, after A is done, KCLO3 and KCL are generated, according to 2KCLO3 = 2KCl + 3O2 and O2 672ml to know the amount of KCLO3 substance, and then the mass of KCL is known from the total mass, so as to know the amount of the substance, and then according to the above formula, the amount of KCL is converted to the amount of KCLO3 consumed in the experiment.

I also participated in the chemistry competition in Shanghai, maybe I can give you some help I finished high school chemistry in my first year of high school, and I used the science class textbooks of Hua II, which is very good, and then I will do some basic competition numbers, such as.

This problem can be calculated using the "conservation of electrons of gain and loss", which is very simple.

First and foremost, it is crucial to clarify the reaction of the individual electrodes

At the cathode, Cu2+2E

Cu on the lead anode.

2cl-2e

Cl2 If the chloride ion reaction is complete, then the anode reacts next.

2h2o-4e-==4h+

O2 is precipitated on the anode.

cu, that is, a total of 2* electrons are obtained.

Theoretically, there should be a chloride ion reaction on the cathode. But!! Pay attention to the relationship between the ratio of the amount of the given substance in the system after the chloride ions, that is to say, at this time only the chlorine gas precipitated on the anode, and the electrons lost.

And there are electrons that are not lost.

However, according to the electrode reaction we have just written, we can know that the remaining electrons will continue to undergo the next step of the electrolysis of water on the anode. In other words, all the remaining electrons come from water, and oxygen is generated

So the total precipitation is out.

Chlorine + oxygen =

Converted to STP, that's it