The problem of putting CuSO4 powder into CuSO4 saturated solution !!

As long as you understand that the crystallization of cuso4 in the presence of water is always in the form of cuso4·5h2o, you can understand this phenomenon. That's what you said (put CuSO4 into a saturated CuSO4 solution, first, CuSO4 absorbs water and becomes CuSO4·5H2O. At the same time, the original CuSO4 solution was already saturated, and some of the water was absorbed by the later addition of CuSO4.

Therefore, not only will the CuSO4 added later become CuSO4·5H2O precipitation, but also the original CuSO4 solution will also precipitate CuSO4·5H2O. )

The reason why CuSO4 can be used as a desiccant is because it absorbs water to produce CuSO4·5H2O

5. Aqueous copper sulfate crystals and copper sulfate saturated solution are not the same concept.

Hehe! Fell into the same trap as I did before!

Because it is already saturated, re-input will not affect the solution, and the injected crystals will exist in the form of crystals at the bottom.

I think so, wait until I discuss it with my classmates.

Strike a balance.

You can check out a similar question here:

Got it?

Repeat until balanced.

The blue crystal is copper sulfate pentahydrate (CuSO4--5H2O).

The above crystals can only appear when all the copper sulfate in the solution is saturated, so your answer is that the crystals are missing and some water is out, and the total amount is not right. Rather, it should be calculated in terms of precipitated crystals.

Because the precipitation is CuSO4*5H2O crystals, not anhydrous copper sulfate.

n-m is part of the original saturated solution. n 90 250 is the water coming out of the original saturated solution. (N160 250)-M is the mass of CuSO4 precipitated from solution.

So the concentration of the original saturated solution, 100%=%.

s=[( n×160\250）-m]÷（n×90\250）]×100=100（16n-25m)/9n

After the crystals are precipitated, the original solution is a saturated solution.

In other words, the mass fraction of CuSO4 in CuSO4*5H2O is higher than that of copper sulfate saturated solution. It is because of the addition of anhydrous copper sulfate that CuSO4*5H2O is precipitated.

A compared with the pathway, the reaction of copper and concentrated sulfuric acid will generate sulfur dioxide gas to pollute the air, the advantages of the pathway: the preparation of the same quality of bile alum requires less sulfuric acid, and the pathway 2 is non-polluting gas production, which better reflects the idea of green chemistry, so a is correct; b The equation for the reaction between glucose and copper hydroxide is: C6H12O6+2Cu(OH)2=C6H12O7+Cu2O+2H2O, so Y can be glucose, and sucrose is a non-reducing sugar that cannot react with iron hydroxide to form cuprous oxide, so B is wrong; c．?

Copper sulfate decomposes to form Cu2O and oxygen, SO3, SO2, according to 2CuSO4 Cu2O+SO2 +SO3 +O2, X may be a mixed gas of O2, SO2 and SO3, if it is only sulfur dioxide and sulfur trioxide, the valency of copper and oxygen elements only decreases, and no element valency increases, so C is wrong; d The CuSO4 solution was evaporated, concentrated, cooled and crystallized, and filtered and washed to obtain bile alum crystals, so d was correct; Therefore, AD is chosen

Reaction: CuSO4 + 5H2O = CuSO4·5H2O

The reaction consumes water, there are blue crystals precipitated in the solution, the solute and solvent mass are reduced, but the solution is still saturated, and the solute mass fraction remains the same.

The precipitation is copper sulfate crystal CuSO4·5H2O, not anhydrous copper sulfate.

Precipitate copper sulfate crystals, which will not react with water and are supersaturated.

The value range of m is (c18 m 128).

With x grams of copper sulfate precipitated, then.

32-x）/（m-9x/16）=1/4

m=128-55x/16

When x 0, m 128

When x 32, m 18

After the crystals are precipitated in the solution, it is still a saturated solution, and the quality of the precipitated crystals can be considered to come from two parts, one part is anhydrous copper sulfate, and the other part is a saturated solution, and the solute in it is x g. Then the mass of the crystal can be expressed as. In addition, the crystal is copper sulfate pentahydrate, and its mass can be expressed as (x+.

These two expressions are equal and can be solved for x= g and the crystal mass is .

Dizzy... This is the molar mass ratio... Abandon books for a year, brother can still remember...

Why. I count out that it is...

25 The solubility of Cuso4 is S grams, and the saturated CuSO4 solution is added to S grams of anhydrous powder CuSO4, so the quality of water that should be added to the original solution is.

b. 100 grams.

Chemical problem: The solubility of CuSO4 is 25g at t degree, add anhydrous CuSO4 powder to 100g of CuSO4 saturated solution at t degree, if the temperature is unchanged, CuSO4 + 5H2O = CuSO4 5H2O160 90 250

x y z first calculates the mass percentage of the saturated solution when t degree: 25 (100+25)*100%=20%.

Therefore, the original 100 grams of saturated solution contains 20 grams of solute.

So the total cuso4 is 20+

160:250=x:z -- just solve it z=

If x grams of copper sulfate pentahydrate can be precipitated, then there is.

x = g "How many g of CuSO4 crystal water can be precipitated?"."What does that mean? Do you want to calculate the mass of the crystalline water in the resulting anhydrous copper sulphate crystals? If yes, calculate grams like this.