Senior 1 Chemistry Calculations. Ask for help, high school chemistry calculations

Option B solution: The amount of the substance with Naalo2 is Xmolalo2- +H+ Al(OH)3

amol amol a mol

At this time, there is ALO2- (X-A)mol left in the solution, and then HCl is added, and ALO2- is considered to be converted into Al(OH)3ALO2- +H+ AL(OH)3

X-a mol X-A mol X-A mol At this time, the total amount of Al(OH)3 is X-A+A=XmolAL(OH)3 dissolved.

al(oh)3+3h+~al3+

xmol 3xmol

Therefore, the total dosage of hydrochloric acid in the second step is x-a+3x=c

x=(a+c)/4 mol

Answer B because.

In Naalo2 solution, after HCl is introduced, the final precipitate is dissolved, so the reaction equation is as follows.

It is explained that only the following reaction occurred

Naalo2+4HCl=NaCl+AlCl3+2H2OHCL is the participating reaction (A +C)mo l to find Naalo2 is (A+C) 4

First by. Naalo2+HCl+H2O=Al(OH)3+NaClAL(OH)3+3HCl=AlCl3+3H2OWe can write the total reaction.

Naalo2+4HCl=AlCl3+2H2O means that 1mol Naalo2 can consume 4mol HCl gas, and we know from the title that the total reaction HCl is A+Cmol, so the amount of Naalo2 is (A+C) 4 mol

Ionic equation: AlO2- +4H+ +=Al3+ +2H2O. So a total of A+CMOLHCI is introduced, so it is A+CMOL hydrogen ion, so it is (A+C) 4molalo2- (metaaluminate ion) so it is A+C 4molNaalo2

I hope mine is helpful to you, don't ask me, feel free to ask for your answers

AlCl3+3NaOH = Al(OH)3+3NaClAL(OH)3+ NaOH = Naalo2+ 2H2O, the white precipitate is Al(OH)3

1.Only when the first step reaction occurs:

n(naoh)=3n(al(oh)3)=3m/m(al(oh)3)=3*

Therefore v(naoh)=

2.Generate naalo2:

N(NaOH)=3N(AlCl3) = N(NaOH)=N(NaOH2)=N(AlOH)=N(Al(Oh)3) Total-N(Al(OH)3) reaction =

So v(naoh)=

Let's talk about ideas.

30ml of 1mol L of aluminum chloride solution has AlCl3The white precipitate that produces the precipitate is Al(OH)3

Then write the following equation, 3NaOH+ALCL3=AL(OH)3+3NACl (1).

al(oh)3+naoh=naalo2+2h2o（2）

When NAOH is insufficient, only reaction 1 occurs

When the precipitate is Al(OH)3 and AlCl3 in the solution, the amount of NaOH and the volume of the solution are calculated according to equation (1).

In NAOH sufficiently.

Precipitates Al(OH)3 but becomes Naalo2 in solution The two equations occur at the same time, and according to equation (1), AlCl3 reacts completely to form Al(OH)3Therefore, according to (2) to dissolve, calculate the amount of NaOH substance used twice.

Solution volume: Solution volume.

AlCl3 + 3NaOH = Al(OH)3 + 3NaCl (NaOH drops into AlCl3).

AlCl3 + 4NaOH = Naalo2 + 3NaCl + 2H2O (AlCl3 drops into NaOH).

There are two cases.

Determine the reaction formula: Al3+ Al(OH)3 precipitation Al(OH)4 - assuming that the amount of NaOH is Xg, 1) has not completely reacted, and there are ions in the system, calculated according to Al3+ 3NaOH Al(OH)3 precipitation.

x: Find x=3*( 40= v=( 4=

2) The reaction has been completely completed, but the excess NaOH dissolves part of the precipitate.

The amount of complete precipitation is calculated first: Al3+ 3NaOH Al(OH)3 precipitation.

The total amount of precipitation is but in fact there is only precipitation, so there is precipitation dissolution, according to al(OH)3 precipitation al(OH)4- is 1; 1 relationship, so the more involved in the reaction, then there is a total reaction.

Calculation v=Done.

Calculations in chemistry are arguably the easiest.

Because the amount of computation is not large, and it is all an integer.

First, you need to be familiar with the mass fraction of commonly used elements, and preferably the mass fraction of commonly used compounds can be written without calculation, for example, the mass fraction of sulfuric acid is 98

Second, the equation should be written correctly.

And thirdly, it's the columnar formula, just like the teacher said.

These are the most basic things you have to master.

Upstairs mentioned the difference method, which needs to be understood, and many students can't understand that solubility is to calculate some percentages, which is simple.

Learn chemistry well, you must be familiar with the periodic table.

The difficulty of the first year of high school is not the calculation problem, but the first chapter of redox reaction, as long as you break this chapter, the rest is easy to learn. Chapter 1: The gains and losses of electronics are handed over to you a mantra.

Rise and loss of oxygen, increase of valency, loss of electrons, be oxidized and reduced by the reducing agent, decrease in valency, get electrons, be reduced The mantra of oxidant was created by the little brother.

If you are about to be in the first year of high school, it is recommended that you review Chapter 8 of Chemistry and the final schedule (dissolution table).

I hope it will help you learn chemistry.

Difference method and solubility calculations.

If you do more, you will get used to it.

When I was learning at that time, I felt that equations were much more difficult than in junior high school, and it was better to memorize them if you had to learn trimming. Of course, as you learn more, this will become very simple in chemistry. However, many rare chemistry questions are simple reaction combinations, so I recommend that you learn the basics.

High school chemistry is very important as long as you get in. Simple.

Hope it works for you, hehe.

When it comes to using mol, many people will not be used to it, but that's all.

There are x grams of sodium sulfite, which reacts with barium chloride to produce y g precipitate.

Then there are grams of sodium sulfate, which can react with barium chloride to produce 45-y grams of precipitate.

na2so3---baso3

x---y217x=126y

na2so4---baso4

Solution x = grams.

Mass fraction of sodium sulfite =

na2so3---so2

z = If these deteriorated sodium sulfites are added to a sufficient amount of hydrochloric acid, the gas can be obtained in liters under standard conditions.

Step-by-step approach (think of s as gaseous).

2h2s+o2=2s+2h2o

112l 56l 112l

The remaining oxygen is then reacted with sulfur.

s + o2 == so21 1 1(

This method of generating and fully combusting H2S gas, accounting for the total volume of H2S, is also the easiest to think of, and it is discussed directly from the reaction process.

Let oxygen first oxidize hydrogen sulfide to sulfur element, and then oxidize sulfur element to sulfur dioxide.

That is, the reaction is carried out step by step.

The first reaction is to oxidize all the hydrogen sulfide to sulfur and there is oxygen left, so the first reaction should be calculated according to the amount of hydrogen sulfide substance.

The second reaction is the remaining oxygen to oxidize sulfur, if this is a sufficient amount of oxygen or an excess, the hydrogen sulfide is equivalent to the complete combustion, so at this time the oxygen must be insufficient, so it should be calculated according to the amount of remaining oxygen.

Ignite Ignite.

2h2s+o2*****2s+2h2o s+o2*****=so2

112l 56l 112l

So in the end, only the generated part of hydrogen sulfide is completely burned, and the rest is converted into sulfur element, which is not completely burned.

So by ignition.

2h2s+3o2*****==2so2+2h20

So some of the hydrogen sulfide is completely burned, so the hydrogen sulfide that is completely burned is.

3NO2 + H2O==2HNO3 + No V3 21 3 2 9, so the water rises at a height of 2 9 in the tube

Let the amount of NO2 be amol, then the amount of Hno3 will be 2A 3mol, and the volume of the gas will be reduced to 2A 3mol.

Therefore, the amount of substance of the solution obtained in the test tube is 1

Twenty ninths of a rise, the mass concentration of the substance is 1 mol l

3no2 + h2o==2hno3 + no

10 3 x to obtain x = 10 9ml, assuming the original 20 3mlNO, so the total volume of NO is 70 9ml. Then the water rises to 10-70 9 = 20 9ml of the total volume 2 9 The substance produced is nitric acid, and if it is the standard condition, then the concentration can be calculated.

CM=1000 W [C is the concentration of the substance, in mol L; m is the molar mass of the substance, in g mol; is the density, in g cm3; w is the mass fraction of the substance, and the unit is g].

According to this formula, m=1000 W C

qw=(ma+nb)/(m+n-p)

So simplify the answer to get 1000q(am+bn) c(m+n-p).

There is no need to consider the problem of overabundance in this question, obviously it is impossible to have insufficient water.

Your algorithm is also right, as long as you put m=

The amount of na is also equal to no longer need to be multiplied by 2, and it is already all. Multiplying by 2 is not right.

The following relative molecular mass is calculated based on the balance of the amount of Na substance, in the reaction the amount of Na substance is constant, one part Na produces one NaOH

You're right to do that.

n=v=