The chemistry questions of the third year of junior high school are calculated by the calculation me

Application Problem 1: Solution: Let the mass of copper oxide be x

h2+cuo=cu+h2o

x 3g80/64=x/3g

x = A: The mass of copper oxide required is.

Application Problem 2: Solution: Let the quality of the water required be x

The mass that requires oxygen is y

Power. 2h2o====h2↑+o2↑

x 100kg y

34/2=x/100kg

x=1700kg

2/32=100kg/y

y=1600kg

A: It takes 1,700 kg of water to be consumed, and at the same time the mass of oxygen is 1,600 kg.

Application Problem 3: Solution: Let the mass of hydrogen that can be produced be x

The quality of zinc chloride is y

zn+2hcl=h2+zncl2

x y65/2=

x≈65/136=

y A: The mass of hydrogen is grams, and zinc chloride is grams.

Is your second question wrong, why do you still ask for the quality of hydrogen when you have already told you about the quality of hydrogen? I wondered if it should be the quality of the oxygen that I was looking for, so I wrote it down.

Application Question 1: Suppose that the mass of copper oxide required is x grams.

H2+Cuo==(Heating) Cu+H2O

x grams 3 grams.

The columnable equation is: x (64+16)=3 64

x = grams. Answer: If 3 grams of copper are made, the mass of copper oxide required is grams.

Application Question 2: Suppose that the mass of water consumed is x kilograms.

2H2O== (energized) 2H2+O2

x 100 kg.

The columnable equation is: x 2*(1*2+16)=100 (2*1*2)x=900 kg.

A: If this oxygen is produced by electrolysis of water, the mass of water required is 900 kilograms.

Application Problem 3: Set: H2 x grams, ZnCl2 Y grams can be prepared.

zn+2hcl==h2+zncl2

The equation for g x y is: g 65=x 1*2

x = grams. y = grams.

Answer: The mass of hydrogen and zinc chloride that can be produced is grams and grams respectively.

Write out the chemical reaction equation, according to the calculation method and calculation law of the chemical reaction equation, you can calculate:

To make 3 grams of copper, the quality of copper oxide is required.

The mass of water to be consumed is and the mass of hydrogen to be obtained is the mass of hydrogen that can be produced and the quality of zinc chloride is respectively.

grams of copper oxide. Kilograms of water, kilograms of hydrogen. The mass of hydrogen is grams, and the mass of zinc chloride is grams.

Do the math. It's very simple, first write the chemical reaction equation, set the unknown quantity, then substitute the correlation equation, and calculate the proportional formula.

2010 Taizhou, Zhejiang) 38Hydrogen sulfide (H2S) is a deadly poisonous gas with the smell of rotten eggs, it is denser than air, and it is soluble in water to form hydrosulfuric acid, which has the general properties of acid. Solid ferrous sulfide (FeS) and dilute sulfuric acid are commonly used in the laboratory to prepare H2S at room temperature, and FeS04 is generated at the same time.

1) In the diagram on the right, the diagram should be selected as the device for producing H2S.

2) When collecting H2S gas, it should enter through the conduit in the C device, and the treatment method for the exhaust gas generated.

Be:. 3) How many grams of 20% sulfuric acid solution are theoretically needed to produce grams of hydrogen sulfide gas in the laboratory?

38．(1)b

2) A Absorption with water (or absorption with alkaline solution, such as sodium hydroxide or lime water, etc.) (3) Solution 1: Let the mass of 20% sulfuric acid solution be x grams.

H2S04 +FES = H2S ++FES0420% x grams.

A, a 20% sulfuric acid solution is required with a mass of 98 grams.

cacl2 + na2co3====caco3 + 2naclx= y= z=

Table salt is mixed with the mass of calcium chloride:

Mass fraction of solute in sodium carbonate solution:

Mass fraction of solute: (

Solution: (1) The precipitate mass must contain calcium carbonate.

Calcium ion = calcium chloride.

Mass of calcium chloride =

2) The number of moles of calcium ions = the number of moles of carbonate = the number of moles of sodium carbonate = the mass of sodium carbonate =

Mass fraction: There is original sodium chloride in the solution New sodium chloride added to water Original water:

So. Mass score ( + (173+ Do your own thing.)

1: Let calcium chloride be x 2: Let sodium carbonate be y

cacl2+na2co3=caco3↓+2nacl cacl2+na2co3=caco3↓+2nacl

x 1g y 1g

x 1g y 1g

x= y=w%= ——

53g w%=2%

3: Let the generation of sodium chloride be Z

cacl2+na2co3=caco3↓+2nacl1g z

1g zz=

mnacl=

m = 25 + 173 + 53-1

w%=——

Let the mass of CaCl2 in the salt be X, the mass of NaCl formed by the reaction is Y, and the mass of CaCO3 precipitate is Z.

cacl2+na2co3=2nacl+caco3↓111 106 117 100

x 100g* y z111/x=106/

x=y=z=10g

The mass fraction of CaCl2 in table salt is:

It turns out that the mass of NaCl in the mixture is:

The mass of the NaCl solute after the reaction is:

The mass of the NaCl solution after the reaction is:

The mass fraction of NaCl in the detected solution is:

Answer: (1) The mass fraction of CaCl2 in table salt is:

2) The mass fraction of NaCl in the detected solution is:

Analysis: (1) In order to find the mass fraction of CaCl2 in well salt, the mass of CaCl2 in well salt must be calculated according to the chemical equation, and then the mass fraction of CaCl2 in well salt can be calculated according to the mass fraction formula.

2) To know the mass fraction of NaCl in the solution after detection, it is necessary to know the mass of NaCl and the mass of the solution after detection, and the mass of NaCl can be calculated according to the chemical equation, and the mass of the solution is the total mass - the mass of the precipitate, which can be calculated according to the formula of solute mass fraction

Solution: (1) Let the mass of CaCl2 be X, the mass of NaCl be Y, and the mass of CaCO3 be Z

The mass fraction of Na2CO3 in Na2CO3 solution is 100g

cacl2+na2co3=caco3↓+2nacl

x z y111 106 =x ，106 117 = y ，106 100 = z

The solution yields x= y= z=10g

The mass fraction of CaCl2 in well salt was: 100%.

100g+ =

Answer: The mass fraction of CaCl2 in the well salt is and the mass fraction of NaCl in the detected solution is.

Add the Na2CO3 solution, so Na2CO3 = , Y by the reaction formula Na2CO3 + CaCl2 = CaCO3 + 2NaCl

So the mass of CaCl2 = so its mass fraction =

2) From the equation, the mass of NaCl = is obtained, so its mass fraction is =

Question 1:

1)m=ρν=

280g×50%=m×20%；m = 700g 700g of dilute solution with a solute mass fraction of 20% can be prepared.

na2co3+ba(oh)2====2naoh+baco3↓

x= y=8g z= \

NaOH in the mixture

The mass of the substance Na2CO3 participating in the reaction in caustic soda is grams.

The purity of this caustic soda is: (4g) 100%=

The total mass of NAOH is: 4G+8G=12G; The mass of the solution is:

The mass fraction of solute in the resulting solution after the reaction is: (12g 120g) 100% = 10%.

Question 2: I can't see the picture clearly.

It is possible to prepare a dilute solution with a solute mass fraction of 20% x grams.

Then: 200*

x = 700 grams.

A must be a base, and a can be introduced as ba(oh)2

So the mass of naco3=

4 NAOH = 8g generated

8g/(

1)m=ρν=

280g×50%=m×20%；m = 700g 700g of dilute solution with a solute mass fraction of 20% can be prepared.

na2co3+ba(oh)2====2naoh+baco3↓

x= y=8g z= \

NaOH in the mixture

The mass of the substance Na2CO3 participating in the reaction in caustic soda is grams.

The purity of this caustic soda is: (4g) 100%=

The total mass of NAOH is: 4G+8G=12G; The mass of the solution is:

The mass fraction of solute in the resulting solution after the reaction is: (12g 120g) 100% = 10%.

Question 2: I can't see the picture clearly.

The solute mass fraction of 20% dilute solution was set to be X grams.

Then: 200 * x = 700 grams of a base, and a is ba(oh)2

Mass of mnaco3 =

4 NAOH=8G8G (

1.(1) It is possible to prepare a dilute solution with a solute mass fraction of 20% x grams.

Then: 200*

x = 700 grams.

2) Don't understand the meaning?

2.Can't see the picture? Write it more clearly.

What a master! Awesome!