The following groups of reactions have no white precipitates in the final solution, each of which is

A. Excessive ammonia water is added to the ALCL3 solution.

Al(OH)3 precipitate is generated, which is insoluble in excess ammonia.

B. Excessive KOH solution was added to Al2(SO4)3 solution.

Al(OH)3 precipitates are generated, dissolved in excess KOH, and finally Kalo2 is formed without precipitates.

c. Excessive ammonia water is added to the MgCl2 solution.

Mg(OH)2 precipitate is generated, which is insoluble in excess ammonia.

D. Excessive KOH solution was added to the MgSO4 solution.

Generate Mg(OH)2 precipitate, insoluble in excess KOH option B

A: AlCl3+3NH3·H2O==AL(OH)3 +3NH4Cl aluminum hydroxide is insoluble in ammonia (weak alkali).

B: Al2(SO4)3+6KOH==3K2SO4+2AL(OH)3 Aluminum hydroxide dissolved in potassium hydroxide solution (strong alkali).

KOH+AL(OH)3==KALO2+2H2O has no precipitation in the end.

C: MgCl2+NH3·H2O--Mg(OH)2 +NH4Cl This is not matched.

D: MgSO4+2KOH==Mg(OH)2+K2SO4 magnesium hydroxide is insoluble in potassium hydroxide solution. Hope.

D. Excessive KOH solution was added to the MgSO4 solution.

a、alcl3

Excess ammonia was added to the solution, and the reaction was AlCl3

3nh3h2

o=ai（oh）3

3NH4Cl, the resulting white precipitate is insoluble in excess of ammonia, so A is correct.

b、al2so4

Excess KOH solution was added to the solution, and the reaction was Al3+

3oh-ai（oh）3，al（oh）3

oh-alo2

2h2o, so b is wrong;

c、fecl2

Excess ammonia was added to the solution, and the reaction was Fe2+

2nh3h2

o=fe（oh）2

2nh44fe（oh）2o22h2

o=4fe（oh）3

Therefore c is wrong; d、mgso4

Excess KOH solution was added to the solution, and the reaction was Mg2+

2OH-mg(OH)2,MG(OH)2 is insoluble in excess KOH solution, so D is correct;

Therefore, AD is chosen

The solution of pH=1 is acidic, and the Mg(OH)2 produced in B is soluble in the acidic solution; CaCO3 in C and Baso3 (barium sulfite) in D are weak salts, and both CO2 and SO2 gas cledges are produced in acidic solutions. The violent laughter in A is AGCL, which is insoluble in acidic solution, so the answer is A

, K2CO3 because the solution is acidic at pH=1, under acidic conditions, CaCO3 will dissolve.

Na2SO3 is the same as Sun Yuzhen's letter, because the pH=1 solution is crude acidic, under acidic conditions, Baso3 will dissolve.

pH=1 solution is strongly acidic, i.e., the concentration of hydrogen ions is larger. Therefore, the precipitate of BCD will be dissolved.

The two undergo redox reactions to form Fe(OH)3

The reaction equation is 2na2o22h2

o=4naoh+o2，fecl2

2naoh=fe（oh）2

2nacl；4fe（oh）2o22h2

o=4fe（oh）3

Therefore, in the end, there is a reddish-brown nuclear friend precipitation, so it is wrong;

Excess Ba(OH)2

The solution and alum solution are mixed, and the two react to form a barium sulfate white precipitate, and the ion equation is Al3+2SO4

2ba2+4oh-

2baso4

alo22h2

o, there is a white barium sulfate precipitate in the end, so it is correct;

A small amount of Ca(OH)2

Excessive amount of NaHCO3 is administered

In solution, the two react to form white precipitated calcium carbonate, Ca(OH)22NaHCO3

caco32h2o+na2

CO3 calcium carbonate is a white precipitate, so there is a white precipitate generated, so it is correct;

na2sio3

Excess CO2 is introduced into the solution

The two react to form white insoluble silicic acid, and the reaction equation is Na2SiO32H2

o+2co2

h2sio3

2nahco3

Therefore, there is a white precipitate generated, so it is correct;

When the carbon dioxide is excessive, the calcium carbonate is converted into soluble calcium bicarbonate, so there is no precipitation in the end.

Therefore, C

A. Although MgSO4 and NaOH can react to form magnesium hydroxide precipitate, but the excess sulfuric acid can dissolve the precipitate, then there is no precipitate in the end, so A is wrong;

B. Because CuCl2 and Koh can react to form copper hydroxide precipitate, and excess KCL cannot react with the precipitation, then the final precipitation is generated, so B is correct;

C. Although CaCl2 and Na2 CO3 can react to form calcium carbonate precipitate, if the hydrochloric acid is excessive, it can dissolve the precipitate, so there is no precipitate formation in the end, so C is wrong;

D. The substances in this group only have the neutralization reaction of NaOH and HNO3, and there is water generation, but no precipitation is generated, so D is wrong;

Therefore, choose B

Select C magnesium sulfate and barium hydroxide react to form two precipitates of magnesium hydroxide and barium sulfate, and only water is left after filtration.

D test question analysis: a, can react with ammonia to produce white precipitation is not necessarily aluminum ions, may also be magnesium ions, etc., a is incorrect; b. Add agno3

solution, which produces a white precipitate, which is not necessarily silver chloride, but may also be silver carbonate, etc., so it cannot be proved that Cl- is present in the original solution

b is incorrect; c. Add hydrochloric acidified BaCl2

solution, which produces a white precipitate, which is not necessarily barium sulfate, but may also be silver chloride, and therefore does not prove the presence of SO4 in the original solution

c is incorrect; D. Add NaOH solution and heat to produce a gas that turns the wet red litmus paper blue, which must be ammonia, so it can be proved that NH4 is present in the original solution

D is correct, the answer is D.

Only BaSO4 white precipitate was generated in A.

In B, Na first reacts with water to generate hydrogen and sodium hydroxide, and sodium hydroxide reacts with MgCl2 to form a white precipitate of magnesium hydroxide.

Aluminum and excess sodium hydroxide in C react to form sodium metaaluminate and hydrogen, 2al+2h2o +2naoh=2naalo2+3h2, and there is no precipitation cover to produce a physical mask.

Neither gas nor precipitation is produced in d.

So the answer is b

C A only produces sock bark baso4 precipitate without gas.

b only produces hydrogen agitation without precipitation.

d. The neutralization reaction has neither gas nor precipitation.

The temperature of A and D is higher than that of B and C, and the reaction rate of B and C is smaller.

Compared with A and D, the concentration is larger, the reaction rate is larger, and the white precipitate can be seen first

So choose A