Ask a junior high school chemistry question in detail

a + b + c ——d

The first reaction is 15-m, 15-n, 15, 30, two reactions, and 15-15-x 25y

The upper and lower corresponds proportionally to y=50 m=6

The mass before and after the reaction is unchanged 15-x=y-15-25 x=515-n=20-15+m-15 n=9

m:n=6:9=2:3

m:n=3:2

Problem solving process: from the first time C reacts to 15 (because the first 15, the mass of C after the reaction becomes 0), and 30 D is generated

It is concluded that in this reaction, c:d=15:30

In the same way, the second increase in the c of 10 is also the reaction (the increase of the c of 10 becomes 0 after the reaction, indicating that the c participating in the reaction is 25), using the above proportional relationship.

c:d=15:30=25：y

Y = 50 is obtained by using the conservation of mass: a + b + c = d participating in the reaction, where the mass becomes 0 after the end of the a reaction, indicating that a = 15 participating in the reaction is substituted into the above equation 15 + x + 25 = 50

X=10 is obtained, and the mass of b in the first reaction can be deduced by the ratio of b:c=10:25.

b:c=10:25=n:15

n=6 is obtained, and the mass of a in the first reaction can be deduced by the conservation of mass.

m+6+15=30

m=9, so m:n=3:2

Substance a, b, c d

Mass after the first reaction (g) m n 0 30

Mass (g) after the second reaction 0 x 0 y

a+b+c==d

According to the law of conservation of mass, for a, c.

15-m):15=15:25 m=9g

The A participating in the reaction for the first time is 15-9=6g, C is 15g, and the resulting D is 30g, so the C participating in the reaction is 30-15-6=9g n=15-9=6g

m:n=9:6=3:2

Na2CO3 + CaCl2 = CaCO3 (down arrow) + 2NaCl, one Na2CO3 produces one Ca2CO3

The precipitate is CaCO3, 3 g, M Na2CO3 = 106 (Na2CO3) 100 (CaCO3) * 3 =

The precipitate is CaCO3, where the mass of element C is 3*12 100=

So the mass of Na2CO3 is:

The mass fraction of Na2CO3 is:

The mass of carbon dioxide produced is 20g+70g+

na2co3+2hcl=2nacl+h2o+co2x

106/x=44/

x = mass of sodium chloride in the original solid.

This is first and foremost a conservation of mass.

It can be known that the mass of carbon dioxide produced is: 20+70+ and the ratio of sodium carbonate to carbon dioxide is 1:1

So, the quality of sodium carbonate is:

So sodium chloride is.

Hope this helps you.

over~~

Through the mass difference, we know that less is the ** of carbon dioxide, and the appropriate amount of carbon dioxide is 44 and the corresponding appropriate amount of sodium carbonate is 90, so the corresponding ratio is sodium carbonate 9g, so the original mass of sodium chloride is 11g

Choose Ca, Fe(OH)3+HCl.

B, CO2+Ca(OH)2---CaCO3C,Cu metal activity is weaker than H, so it cannot be directly added from Cu HCl to CuClD, MgCl2+K2SO4.

According to the information provided in the title, it is known that copper hydroxide is first heated and decomposed, which is the AB segment in the image, and the solid mass of the AB segment is reduced, and the reduced mass is the mass of decomposition to produce water.

It can then be calculated by the chemical equation: Cu(OH)2====CuO+H2O Losing water requires Cu(Ho)2 mass to. Calculated a=

From the diagram, we can see that the mass of the CD segment decreases because Cu2CO3===CuO+CO2 and the mass of CO2 decreases to, and only the CuCO3 of the molecule is required to produce CO2 by calculation, i.e., B= so A:B=2:1

From the information, it can be seen that the water is lost first and then CO2

As you can see from the diagram, the lost water has so there is water.

According to the formula Cu(OH)2=(Heating)=CuO+H2O, Cu(OH)2 is required

Look at the lost CO2 and there is CO2 again

According to the formula CuCO3 = (heating) = Cuo + CO2, so the corresponding CuCO3 has.

So their coefficient is 2:1

1.According to the title: Let the chemistry of this oxide be R2O3 (because oxygen is -2 valence and this metal is +3, so the chemistry is so).

According to the known conditions, the relative atomic mass of this metal r is x (according to the law of conservation of mass, the mass of the solution after the reaction is 16g+184g=200g, so the mass of the solute after the reaction is 200*20%=40g).

According to the problem, the equation is: R2O3+3H2SO4 = R2(SO4)3+ 3H2O

2x +48 2x+288

According to the known two substances.

Exactly 16g of full reaction 40g

Proportional formula (2x+288) (2x+48) = 40g 16g x = 56

2.According to the title, the mass of sulfuric acid consumed is y

r2o3+3h2so4 = r2(so4)3+ 3h2o

The proportional formula of y 40g is 294 400 = y 40g y =

According to the formula, the mass fraction of solute = (mass of solute mass of solution)*100%.

So: Mass fraction of solute = (

The chemical formula of the combination of trivalent metals and negative divalent oxygen elements is generally R2O3, which is one of the things to consider.

Secondly, the atomic weight is required, through the known conditions given in this question, it is generally possible to think of using the proportional formula under the equation to find, then the proportional formula is still a number, that is, the mass of pure sulfuric acid, the quality of pure sulfuric acid is required to use 100 molecules 20, and the mass of sulfuric acid r can be found with 20% (note the writing of the chemical formula of sulfuric acid r), considering this, we use a more direct proportional formula, that is, the real mass of oxidation r and sulfuric acid r is compared with their relative masses respectively, where the atomic weight of r is x, It should be a one-dimensional equation, and the solution can be done.

Once the atomic weight of r is known, the true mass of pure sulfuric acid can be calculated, so that the solute mass fraction is equal to the solute mass divided by the mass of the solution and multiplied by 100%, and the remaining problems should not be big.

No thanks.