Senior one chemistry, kneeling and begging for a master s answer, grateful!

NAOH is insufficient.

al3+--3oh---al(oh)3

x x = volume of NaOH =

NAOH slightly overdosed.

al3+--3oh---al(oh)3 al(oh)3---naoh---naalo2

A total of Al(OH)3 is produced, the second reaction reacts , and the remaining NaOH total consumption volume =

The Al2(SO4)3 solution, containing 2* molar Al3+, can generate molar Al(OH)3 precipitate with a mass of g.

Now there are only grams (moles) of precipitates, and there are two possibilities: one is that the amount of NaOH is small, which is only enough to generate gram precipitates, and moles of NaOH are needed, and the volume of the solution = liters = 150 ml; The second is an excess of NaOH, the gram precipitate dissolves the gram precipitate, and molar NaOH is required, and the volume of the solution = liter = 200 ml.

Of course, OH- and H+ in A cannot coexist in large quantities, and the reaction generates H2OB Since Ca(OH)2 is a microsoluble, Ca2+ and OH- cannot coexist in large quantities, otherwise calcium hydroxide will precipitate and precipitate.

C H+ and CO32- cannot coexist in large quantities, otherwise H2O and CO2D can coexist in large quantities maliciously.

So I believe I was right because I was in my third year of high school and had done countless questions like this.

A H+, Oh-B Ca2+, Oh-, (calcium hydroxide is a microsoluble substance how to coexist in large quantities).

C H+, CO3 2- (3 at the bottom, 2- at the top).

d d is right, don't be too superstitious about the answer.

The right one is D, and the B is wrong! Ca(OH)2 is slightly soluble in water and cannot coexist in large quantities!

Both sulfur ions and sulfite ions are mainly reducible, and can coexist under alkaline conditions, while centering reactions occur under acidic conditions. According to the title, the following reaction occurs after adding sulfuric acid to the solution: SO32- +2S2- +6H+ = 3S +3H2O; (If the reaction happens to be complete, the ratio of the amount and concentration of sulfite and sulfur ions is 1:.)

2) Because of the formation of SO2 gas, it can be seen that there is an excess of sulfite, and the excess sulfite undergoes the following reaction: 2H+ +SO32- = SO2 +H2O generates SO2 gas.

Therefore, the ratio of the concentration of the amount of sulfite to sulfur ion is greater than 1:2;

So the answer is C.

c.The sulfur ion and sulfite are neutralized under acidic conditions: 2S +SO6H =3S +3HO

And because there is also so generation, it means that there are too many sulfite ions in the solution. Therefore, the ratio of the two is greater than 1 2

1.A small amount of solid is taken and put into water, which produces both gas and precipitation.

There must be KhCO3 in gas, only it may produce gas, and it must react with H+ to have gas, so there must be NaHSO4, which can ionize H+, and the precipitation is not analyzed first, and there are more possibilities.

2.Filtration, a small amount of filtrate was added to the AgNO3 solution, and no precipitation was generated.

There must be no CaCl2, so the only cation that can generate precipitated cations is barium ions, so there must be Ba(NO3)2

3.After taking a small amount of filtrate and adding NaOH solution, no precipitation was generated.

Addition of NaOH solution no precipitate and certainly no MgSO4.

Nano3 could not be determined.

NaHSO4 can be seen as a strong monobasic acid.

Therefore, if there is a gas, it can be judged that there must be KhCO3, and another precipitate must be generated.

Since the AGNO3 solution is added, no precipitation is generated, so there are definitely no CL ions.

Add NaOH solution, and there is no precipitation, so there are definitely no mg ions.

Then the precipitation can only be caco3.

Therefore, NaHSO4, KhCO3, and BA(NO3)2 must exist.

CaCl2, MgSO4 certainly don't.

Nano3 has no effect on the above reactions, so it is impossible to be sure.

NaHSO4 can electrically produce hydrogen and sulfate ions in water, and HCO3 combines with H to produce gas. There is precipitation production, and there are BA ions.

There is no precipitation and no chloride ion with silver nitrate.

NaOH is added without precipitation and no mg ions.

There must be no CL ions, and if there were, there would have been precipitation.

Conservation of charge2x+y=3m Conservation of elements x=m (according to r) ; y=2n (according to h),; n=2 (according to the oxygen element), so m=y=4

R3 is obtained by oxidation of R2 electrons.

R2 is a reducing agent, so R3 is an oxidation product.

ab is right, this question has been tested many times, the classic question, which examines the conservation of elements, the conservation of electrons and redox reactions.

According to the conservation of the r element, we know that x=m

According to the conservation of charge, 2x+y=3m is known

According to the conservation of the o element, 2=n is known

According to the conservation of the h element, y=2n is known

According to the above equation, we know that n=2 and y=4

So x=m2x+4=3m

The solution is m=x=4

So the cd is excluded.

The valency of r increases from +2 to +3 and is an oxidation product.

So choose B

First of all, R2+ becomes R3+, then R2+ is the reducing agent, and R3+ is the oxidation product, so as to exclude CD

yh+ +o2 gives nh2o and knows n=2, y=4Since it is an ionic equation, the conservation of charge yields: 2x+4=3m

Actually, don't do this, it's a multiple-choice question, and when y is found, you can exclude b, because here y is a fixed value, and you don't need to be like b.

The correct answer is B

According to the conservation of h, y=4, according to the conservation of charge, 2x+4=3m and x=m, then m=4

r increases from +2 valence before the reaction to +3 valence after the reaction, which is oxidized and is an oxidation product.

From the ionic equation charge conservation it can be concluded that x : y = 1 : 1 from the conservation of r element matter it can be concluded that x = m

From the above 2 conclusions: x = y = m

From the conservation of the o element matter, it is concluded that n = 2

From the conservation of the h element matter and the above conclusions, it is concluded that y = n * 2 = 4 from all the above conclusions: x = y = m = 4, n = 2o2 is an oxidizing agent, r2+ is a reducing agent, r3+ is an oxidation product, h2o is a reduction product, option a is a bit vague, oxide does not know whether it refers to an oxidant or an oxidation product, option b is correct, but it is not complete.

c and d are completely wrong.

Therefore, the answer is reasonable to say A, B is also reasonable, and both AB are reasonable, and the questioner is very unrigorous.

From the equation one oxygen molecule and 4 hydrogen ions get two water molecules, so n=2, y 4, an oxygen molecule loses 4 electrons, and then R2+ changes to R3+ needs to get 4 electrons, so every R2+ needs to get 1 electron, so the valency of m=4 r increases, it should be an oxidation product, so B should be chosen