Chemistry questions on metals in high school, chemistry questions in higher grades

Because the reaction will produce hydrogen, there is a limit to hydrogen (when hydrogen and air are mixed in a ratio of 4%, it will occur during combustion), only high or low than this range will not**, the title says: isolated from the air, so the purity is very high, will not **, safe. Therefore, the purification in the production of hydrogen is this reason.

。。Could it be that there is a layer of oil above the water layer that reacts with water... In this way, the reaction between sodium and water will cause sodium to float to the surface of the water and the oil layer on the water layer does not react with na, so it is safe.

At the same time, because the oil layer can isolate the air and prevent the impurity of the collected gas, it is not possible to prevent the impurity of the collected hydrogen, resulting in **... Since there is no diagram, these are all conjectures...

Because sodium can react with oxygen in addition to water, so in order to ensure the accuracy and reliability of the experiment, it is necessary to exclude other irrelevant variables, so as not to affect the experiment, and this problem is to test the generated gas, so it is necessary to exclude the possibility of reacting with oxygen, so as to ensure that sodium is reacting with water.

I think this kind of topic should be the basic question in the book, read the book carefully, and remember the reactions and things clearly, so that you will know the benefits after the third year of high school.

The density of sodium is greater than that of kerosene, and he will sink below when placed in kerosene, this device has no diagram now, it is estimated that it is a layer of hydrated kerosene, sodium is in the middle, so that there is no air contact, the results are different from the general experiment, first, sodium is not in air contact, that is to say, if there is a violent reaction, sodium will not splash out, and second, sodium and the generated hydrogen will not be oxidized immediately, reducing the reaction exotherm.

If you put sodium directly into water, the density of sodium is greater than that of water, it will float on the surface of the water, it will come into contact with the air and burn, the experimental device you mentioned, because the density of Na is gcm-3, so the sodium will float on the water and sink under the kerosene, so that the sodium only reacts with water and not with air, so that the sodium will not burn, this is the meaning of the answer.

Over, I hope it helps.

A certain amount of FeO, Fe2O3, Fe3O4 mixed powder and a certain amount of aluminum powder are evenly prepared into a thermite and divided into two parts, the first part is completely reacted at high temperature to form alumina and iron, and then the product is fully reacted with a sufficient amount of dilute sulfuric acid.

Fe is all ferrous iron.

The second part directly added a sufficient amount of sodium hydroxide solution to fully react, and FeO, Fe2O3, Fe3O4 did not change.

For this reason, the first part produces less hydrogen.

The answer can be a, c, d

H216O, D216O should be ordinary water and heavy water, heavy hydrogen in heavy water contains one neutron and the atomic weight is 2

1): 36g of ordinary water and 60g of heavy water are 2 moles and 3 moles respectively, so the ratio of the number of atoms they contain is 2:3

2): Hydrogen in ordinary water does not contain neutrons, oxygen contains 8 neutrons, heavy hydrogen in heavy water contains 2 neutrons, oxygen contains 8 neutrons, 2 8:3 10=16:30=8:15

3）：2na+2h2o＝2naoh+h2↑

According to the chemical equation.

2 moles of ordinary water produce 1 mole of hydrogen or 2 grams of hydrogen, and 3 moles of heavy water produce moles of 6 grams of heavy hydrogen, so they react with a sufficient amount of sodium metal respectively to produce a mass ratio of 1:3 for gas

36g of water is 2mol, so h 4mol o 2mol60g heavy water is 3mol so d 6mol o 3mol1The ratio of the number of atoms contained is 4+2 to 6+3=2:32

The number of neutrons contained is more than h without neutrons o contains 8 neutrons d has one neutron.

So it's 8*2:(1+1+8)*3=16:303With the reaction with a sufficient amount of sodium metal, the mass of the gas produced is 1mol of hydrogen than that of water and heavy water is deuterium gas (d2).

So 2:6

First of all, it is clear that the number of atoms can be directly counted as horns, h has no neutrons, d has one neutron, and 16o has 8 seeds.

36GH216O (hereinafter referred to as water) is 2mol, and 60GD216O (hereinafter referred to as weighing water) is 3mol, which can be seen here as a comparison between 2 parts of water and 3 parts of heavy water. So the ratio of atoms to numbers is 2:3.

The ratio of neutrons is 2*(0*2+8): 3*(1*2+8)=16:30=8:

5 。2 parts of water reacts with sodium to form 1 part of hydrogen, and according to 2D2O + 2NA ==2NaOD + D2, 3 parts of heavy water and sodium react to form part of heavy hydrogen (gas), so the mass ratio of the gas produced is 2*1::6=1:

3 Hope mine can help you.

Under the same conditions, the ratio of gas density is equal to the ratio of relative molecular weight, and the density of CO and N2 is the same with the same molecular weight. So CO is arbitrary volume. The key is the volume ratio of CO2 and H2. You can use the criss-cross method.

co2 h2

26:16==13:8 So the volume ratio of CO2 to H2 is 13:8 and CO is arbitrary volume.

Let CO2 have x mol, H2 have ymol, and co has n mol, and under the same conditions, there are equal numbers of molecules in the same volume of gas, so as to obtain 44x ten 2y 28 * (x ten y), obtain 8x 13y, and obtain x:y 13:8.

Since the formula of cos is equal to n2, it is also possible that x:y is 0:0.

To sum up, the volume ratio of the three is 13:8: n, and n is an arbitrary number; Or 0:0:n, n is non-zero.

Add sodium metal to 100 ml mol L of AlCl3 solution to complete the reaction, and just produce a clear solution containing only NaCl and Naalo2, then the mass of sodium metal added is ( ).

According to the conservation of Al and Cl, Naalo2 is mol; NaCl is mol, total n(na) = mol, m(na) = g.

(1) Set x grams of sodium metal.

2na+2h20===2naoh + h2x(g) 36x 46 80x 46 to solve the equation (80x 46) :(100-36x 46)=22:100 (2) from the reaction formula 2na+2h20===2na+ h2, there will be na+, which consumes water.

So it turns out that the water should have, i.e. grams.

From the reaction formula, it is calculated that the overflow of hydrogen gas is immediate.

The mass of the solution is:

Colorless and transparent, indicating that there must be no Cu2+

When Al3+ is generated, it means that the solution is acidic, so there must be H+, so there must be no OH-, HCO3-, and because of the formation of H2, there is no NO3-, because if there is an acid solution of NO3- will not produce H2, so there may be H+, Cl-, Ba2+, Mg2+, or H+, SO42-, Mg2+, Cl-

When alo2- is generated, it means that the solution is alkaline, so there must be OH-, so there must be no Mg2+, H+, Ag+, because it will react with OH-. So there must be Ba2+, and the solution must contain anions and cations, so there may also be oh-, cl-, no3-

First of all, according to the topic, the impossible is excluded.

Solution colorless: then the colored Cu2+ blue ones are excluded.

In the case of a solution, many ions cannot coexist together, such as Ag+ and Cl-, H+ and Oh-, H+ and HCO32-, Ba2+ and SO42-

If Al3+ is required, there must be H+ in the solution, otherwise no gas will be generated, so OH- and HCO32-, as well as NO3-, because if there is an acid solution of NO3-, H2 will not be produced, AG will not be there, and if there is, AG will be replaced, so the possible ions are H+, Cl-, Ba2+, Mg2+, or H+, SO42-, Mg2+, Cl-

When alo2- is generated, it means that the solution is alkaline, so there must be OH-, so there must be no Mg2+, H+, Ag+, because it will react with OH-. So there must be Ba2+, and the solution must contain anions and cations, so there may also be oh-, cl-, no3-

For this kind of question, you should first remember to pay attention to those ion pairs that cannot coexist, and then eliminate them step by step.

Reaction of aluminum with acids and bases:

2al +6hcl ==== 2alcl3+ 3h2↑

2AL + 3H2SO4 (dilute) ==== AL2 (SO4)3 + 3H2

2al + 2naoh + 2h2o ==== 2naalo2+3h2↑

When Al3+ is generated, the solution is strongly acidic, with strong acid, and because it is a colorless and transparent liquid, see if other ions will react in the acid solution to form gas or precipitation, it will be eliminated, and it will not be retained. The colored ones are excluded too.

When ALO2- is generated, it is an alkali solution, and then it depends on whether other ions will react in the alkali solution to form gas or precipitation, which will be eliminated and will not be retained.

1.It is explained in the question that it is a colorless transparent solution, so Cu2+ cannot exist.

2. When Al3+ is generated, it means that the solution is acidic. Therefore, there may be ions with h+Both Mg2+ and Cl- while Ba2+ and SO2- may be present.

So there are two answers. HCO3 - cannot be present because in an acidic solution. NO3 - is also not likely to exist, because in acidic solutions it is strongly oxidizing and hydrogen is not possible.

3.When alo2- is generated, the solution is alkaline. Therefore, there may be ions oh-, cl-, no3- and mg2+, ag+, which will form precipitates, so only ba2+ exists.

This kind of problem examines the reaction between various substances and the physical properties of matter, including chemical properties, so to do this kind of problem well, it is necessary to be proficient in the physical and chemical properties of matter, as well as the equations and reaction conditions of chemical reactions. 、

The first empty because of the generation of aluminum ions so hydrogen ions, no hydroxide ions, colorless so no copper ions, bicarbonate and aluminum will be double hydrolyzed, nitrate will passivate aluminum, the rest are possible, sulfuric acid and barium ions can not coexist, so two sets of answers.

The second space is the same as hydroxide ions, no hydrogen ions, magnesium ions, carbonate ions, and silver ions. Because it reacts with hydroxide, there are barium ions, no sulfate, and the rest is possible.