Help me with a chemistry problem. A chemical calculation problem.

al(oh)

There are two cases:

1. Insufficient NAOH.

AlCl3 + 3NaOH = Al(OH)3 + 3NaClX = Volume: i.e. 180ml

2. If there is too much NaOH, all AlCl3 will generate Al(OH)3 precipitate, and the precipitate will react with NaOH and the remainder.

AlCl3+3NaOH = Al(OH)3+3NaCl remaining precipitate, indicating the reaction.

Al(OH)3+NaOH=NAALO2+2H2OThe total amount of NAOH consumed is .

Volume: , i.e. 340ml

1), a small amount of NaOH and the precipitate is Al(OH)3, then its molar mass is , and the whole Al is, so the volume at this time is; (2) NAOH overdose, 300ml+

Do it yourself, it's not good to play chemical equations on the Internet, and there are two answers, one is a small amount of NAOH, and the other is a little too much NAOH.

3cu + 8hno3 = 3cu(no3)2 + 2no + 4h2o

Consumption, so HNo3 is excessive, and the Cu reaction is complete.

3cu + 8hno3 = 3cu(no3)2 + 2no + 4h2o

1) The volume (standard condition) of the gas generated by the drainage gas collection method is maximum.

2) After the end of the reaction, the amount of hydrogen ions in the resulting solution (the volume change of the solution before and after the reaction is negligible) is the concentration of hydrogen ions (

(1) l o grams; (2)20．6 5％

Since the iron powder is put into the solution, the copper in the solution will be replaced, and 56 grams of iron can replace 64 grams of copper, that is, the mass of the solid will increase, and the mass of the final dry solid is 20 grams, which means that there is still sulfuric acid in the solution, that is, the added cuo is all reacted, and the sulfuric acid and iron powder continue to react, so the mass of the solid does not increase. And 14 sulfuric acid and 8 grams of iron at most, so, iron powder is excessive. In this case, let the mass of cuo be xg.

cuo＋h2so4＝cuso4＋h2o （1）fe＋cuso4＝feso4＋cu （2）（1）+（2）

Cuo Fe H2SO4 Cu H2O 80 grams of copper oxide and 98 grams of sulfuric acid react, 56 grams of iron are consumed to produce 64 grams of copper, and the solid weight gain is 8 grams The whole weight gain: 8 x 80 x 10

Sulfuric acid consumed: 98x 80

fe＋h2so4＝feso4＋h2

56 grams of iron and 98 grams of sulfuric acid react, the solid increment is reduced by 56 grams, according to the topic, the mass of iron powder reacted with (14-98 x 80) sulfuric acid because: x 10, then 56: (x 10) = 98: (14-98 x 80) solution x 10

Assuming that the mass of FeSo4 in the solution is Y grams, it can be seen from the above analysis that the final solute is FeSo4, and all of it is generated by H2SO4.

98 grams of sulfuric acid can produce 152 grams of FeSo4.

98：14＝152：y

Y g mass of the whole solution: 100 10 h2 mass.

The mass fraction of the solute in the solution can be obtained by further comparison.

1. There is gas production.

3、cacl2、hcl

4. It seems that B is not the midpoint of AC.

If so, it's easy to do:

2、caco3 2 hcl cacl2 co2x 156x10% y z=

x = 20 grams.

So the a coordinate is (0,20).

4. Y = grams.

Mass Score = Divide by (156+.)

2kClO3=mNO2, heating = 2kCl+3O2 The mass of oxygen produced is (m-p)g

The mass of the mnO2 does not have to be counted, and the Mno2 is not involved in the reaction.

This has nothing to do with whether or not there is a complete reaction.

Regardless of whether it is completely reactive or not, the mass of mNO2 is ng.

Potassium chlorate, which is not decomposed, is not involved in the reaction and does not affect the calculation.

Imagine, if you still want to add a g of potassium chlorate before the reaction, but you don't add it, and add a g of potassium chlorate after the reaction, will this a affect the calculation? Obviously, it doesn't affect, and a g of potassium chlorate here is equivalent to potassium chlorate that has no reaction.

According to the law of conservation of mass, the mass of oxygen produced is: m g-p g

n g manganese dioxide is used as a catalyst for this reaction, and its mass is not counted.

Simple, manganese dioxide is the catalyst. It does not participate in the reaction, so the oxygen produced is (m-p)g