The existing mixture of Cu and CuO was measured to have a mass fraction of 10 for oxygen

Agree with the solution on the first floor, and I'll talk about my understanding.

The first thing to determine is that in the mixture of cu and cuo, only cuo will react with sulfuric acid, so the mass of cuo in the 16g mixture is required.

Secondly, it is obtained according to the cross method.

The mass fraction of oxygen in Cu was 0, and the mass fraction of oxygen in Cuo was (16 80=2 10).

The mass fraction of oxygen after mixing the two is 10 = (1 10)cu 0 1 10 1

Mixture 1 10

cuo 2/10 1/10 1

According to the cross method, the ratio of Cu and Cuo in the mixture = 1 1, then the mass of Cuo is 16g*50=8g

Finally, according to the reaction equation, the mass of copper sulfate is 16g

The mass fraction of oxygen is 10% and the mass fraction of cuo is 10% (16, 80) = 50%.

16g mixture contains copper oxide 16g * 50% = 8g

cuo+h2so4=cuso4+h2o

8g x80/8g=160/x

x=16g

Since the oxygen content is 10%, then the copper content, that is, calculated by the mol number, is that copper sulfate can be formed by continuous reaction, that is:

1. Because copper oxide is dissolved in dilute salt and lacks acid, it can make the mixture react with excess dilute hydrochloric acid, and then filter, what remains on the filter paper is the copper that has not participated in the reaction, and subtract its mass from the volcanic mass of the original mixture is the mass of Cuo, and then divide by the total mass to obtain the mass fraction of Cuo, the chemical equation Cuo + 2HCl = CuCl2 + H2O;

Second, the mixture is heated in oxygen, so that the copper can fully react with oxygen to form copper oxide, and then weigh the mass after the reaction, subtract the mass of the mixture before the reaction to obtain the mass of oxygen participating in the reaction, and then according to 2cu + o2 = ( 2cuo, the mass of copper participating in the reaction can be calculated, and then the mass of copper oxide and its mass fraction can be calculated. Chemical equation: 2Cu+O2=( 2CuO,2, Determine the mass fraction of Cuo in Cu and Cuo mixture?

For both methods, chemical equations should be attached.

The answer is pretty much the same, the only available reagents are zinc granules, dilute H2SO4, NaCl solution, and concentrated sulfuric acid.

Since the remaining solid mass is equal after adding 120 grams of sulfuric acid solution to Ding and 100 grams of sulfuric acid solution to A, it means that the copper oxide in the sample is reflected. Therefore, for a 20 g sample, there are 8 g of copper oxide. This allows the mass fraction of copper oxide to be determined, which determines 10 grams of copper oxide in a 25-gram sample.

As a result of B, the copper oxide is not completely reacted, and there is still a surplus, indicating that the sulfuric acid in the sulfuric acid solution is fully reflected. According to the chemical equation of the reaction of copper oxide and sulfuric acid, where the copper oxide participating in the reaction is, the corresponding sulfuric acid, 16 grams. So the mass fraction of the solute in the sulfuric acid solution is.

Summary. The relative atomic masses of copper (Cu), sulfur (S), oxygen (O) and hydrogen (H) are 1,1 molecules CuSO4 and 10 respectively. Therefore, the mass ratio of Cu, S, O, and H elements in CuSO4 5H2O = 1 64:

The mass fraction of oxygen in Cu+SO+4 5H2O is.

Hello, the mass fraction of oxygen in Cu+SO+4 5H2O is.

The relative atomic masses of copper (Cu), sulfur (S), oxygen (O) and hydrogen (H) are 1,1 molecules CuSO4 and 10 for 5H2O, respectively. Therefore, the mass ratio of cu, s, o and h yuan mengyin xiaosu in cuso4 5h2o = 1 64: zhizhu 1 32:

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cuo is transposed to cu, and the product loses the part o, that is, 8, a chemical problem cuo + co = cu + co2 hu zheng which m (is the mass of oxygen).

After burning in a closed container, the products are CO2, CO and H2O, after the product passes through concentrated H2SO4, the concentrated H2SO4 is weightless, and after the full reaction of the burning CuO, the Cuo is weightless, and finally through the soda lime, the soda lime is weight-gained, if the same amount of the organic matter and sodium just react Gauge the structure and name of the organic matter and (1) write the structure and name of the organic matter

How to understand the phrase "making cuo weightless" after the full reaction of the scorching cuo.

Because cuprous oxide disproportionates, producing copper and copper oxide, this question can be seen as a mixture of copper and copper oxide.

Let the quantities of cu, cu2o, and cuo be xmol, ymol, and zmol, respectively. Then the total amount of copper contained is: x+2y+z

mol, and the amount of copper elemental substance is: x+y

mol, the amount of the total oxygen element is y+zmol, a part is introduced into a sufficient amount of h2, and the solid mass is reduced, then the mass of the oxygen contained in it is g, that is, there is: y+z=

Then according to: 3Cu + 8Hno3 = 3Cu (NO3) 2 + 2No + 4H2O3mol

n(cu) i.e.: x+y=

The total amount of copper contained in the substance is: the amount of the total amount of nitric acid consumed by x + 2y + z = twice the amount of copper + twice the amount of no gaseous substance =

The concentration of Hno3 added during transformation is:

mol/l。

Solution: Let n(cu)=x in one of the servings

mol，n（cu

o) = ymol, n (cuo) wax shen = z

mol, conservation of electrons and oxygen according to gains and losses, there are:

2x+2y=

3=y+z=

16g/mol

x+2y+z=

That is, the amount of nitric acid that is produced after the reaction is.

then 500 ml of dilute nitric acid, n (hno

Suibu = thus: c(hno=,

No need to explain in detail, just a little trick.

In the three chemical formulas, the ratio of the number of copper atoms to sulfur atoms is the same, both are 1:1, so the mass ratio of copper and sulfur is 64:32=2:1, and the mass fraction of sulfur is known to be x%, and the mass fraction of copper is 2x%. The rest is oxygen 1-3x%.

There are many similar problems, such as ferrous sulfate, a mixture of iron sulfate, the mass fraction of sulfur is known to be a, and the mass fraction of iron is found. (Please answer it yourself, and communicate if you have any questions).

Let's ...

First of all, if you calculate that all iron is FeO, then 28% (56 72) = 36% gives Cuo is 64

When all iron is Fe2O3, then 28% (56*2 (56*2+16*3))=40% gives Cuo is 60%.

The CUO quality score is between 64 and 60, which is only C

The mass of the sample that decreases before and after the reaction is the mass of Cuo, and the mass of the remaining solid is the mass of Cu.

Compared with A, B and C, it can be seen that although the mass of the initial sample is decreasing, the Cuo consumed by the reaction is always 8g, indicating that the sulfuric acid in A and B is always insufficient.

Compared with propylene butylene, it can be seen that although the sulfuric acid used has increased, the quality of the reduced has not changed, indicating that the cuo in butylene is insufficient.

Obviously, propylene is sulfuric acid that happens to react completely with Cuo. It was concluded that 100 g of sulfuric acid solution was exactly completely reacted with 8 g cuo.

The equation is obtained by the relation h2SO4 cuo

The mass fraction of sulfuric acid was 8 (16+64) (2+32+16 4) 100=, and the mass fraction of Cu in the mixture was 1-8 (16+64) 16 20=92%.