Simple Chemistry Calculation Questions for Junior High School Students Find 6 of the easiest chemist

It seems that the quantity of matter was only learned in the third year of junior high school chemistry, and the quantity of matter in other places was only learned in the first year of high school!

This problem can be solved using either the relational method or basic chemical equations.

Solution: (1) The chemical equation is adopted.

nahco3 + hcl ==== nacl + h2o + co2 ↑

20g * w%

84/(20g * w%)=44/

Calculated: w%=84%.

2) Relational method (in fact, it is derived on the basis of chemical equations, the key is to pay attention to the constant mass of the central element C, and use the law of conservation of mass well).

nahco3~~~co2 ↑

20g * w%

84/(20g * w%)=44/

Calculated: w%=84%.

As can be seen from the title, CO2 is generated

NaHCO3+HCl==NaCl+H2O+CO2 x Solution: x=

Quality Score = A: --

Solution: What increases is the mass of CO2. Let the mass of NaHCO3 be ANAHCO3+HCl=NaCl+H2O+CO2A

Solution: a=

20g A: The mass fraction of NaHCO3 in baking powder is 84%.

As can be seen from the question, carbon dioxide is generated.

nahco3+hcl==nacl+h2o+co2↑x

Mass Score =

To increase the mass of carbon dioxide, NaHCO3-CO2-NA2CO3

x mass fraction = = 84%.

The added mass is the mass of carbon dioxide, ie. Then the number of moles of carbon dioxide is due to the formation of 1 mole of carbon dioxide per mole of NaHCO3 reaction (the reaction equation can be written), then there are moles of NaHCO3. The quality score is:

The weight of CO2 is increased. Because NaHCO3+HCl=NaCl+CO2+H2O, sodium bicarbonate can also be calculated as a mass fraction of 84%.

1) It is known that the mass fraction of potassium chloride saturated solution at 30 is 27%, and the solubility of potassium chloride at 30 is found.

Solution: s = solubility.

p = quality and malpractice of this smear score (% removed).

s=100p/（100-p）

100 27) Pickpocketing (100-27) = 37 grams.

Answer: The solubility of potassium chloride at 30 is 37 grams.

2) How many grams of 95% sulfuric acid need to be added to increase the concentration of 50% sulfuric acid to 20%?

Solution: Suppose 95% sulfuric acid is added. x grams.

p% = m fluid m.

50 grams. x) (50g+x).

20% = (g +.)

x) (50g+x) 100%.

g + g + g g. x = 10 grams.

3) 100 grams of sulfuric acid solution with a mass fraction of 98% diluted into a 10% sulfuric acid solution, how much water needs to be added to rent?

Solution: Let the mass of the solution to be diluted be x.

100g 98%=10%*x

x = 980 grams.

m plus water = 980 g - 100 g = 880 g.

There is a marble sample in which the calcium content is 36%, and the mass fraction of calcium carbonate in this marble sample is determined.

Answer] 36% 40% = 90%.

This question is a difficult problem for junior high school students.

Just look for it, and you'll find it next!

In order to determine the mass fraction of calcium carbonate in shells, students from a chemistry interest group took 10 grams of shell samples and reacted them with a sufficient amount of HCl in a beaker after crushing the shells (assuming that the impurities in the shells do not react and are not soluble in water) and measured the total mass of the substances after the reaction decreased in grams, and obtained: (1) the mass of carbon dioxide produced by the reaction was --- grams: (2) the mass fraction of calcium carbonate in the shell sample.

1。A sample of 25 grams of impure potassium chlorate was mixed with manganese dioxide and heated to produce oxygen. If 10% of oxygen is lost during the collection process, only grams of oxygen will be collected. Find the mass fraction of impurities in a 25 g potassium chlorate sample.

Solution: Let the mass of kclo3 in the sample be x g, which is known from the title:

The mass of the reaction to produce O2 is approximately equal to the reaction equation.

mno22kclo3*****==2kcl+3o2↑

The x-columnable equation is 245 x=96

The solution is: x = so the mass fraction of the impurity is (

A: The mass fraction of impurities is 2%.

2.8 grams of an iron oxide is passed into a sufficient amount of hydrogen, and heated to fully react to obtain gram water, and the chemical formula of this oxide is obtained.

Solution: Let the chemical formula of the oxide be Feaob, then the equation for the reaction with hydrogen is:

Heating Feaob+BH2====AFE+BH2O

56a+16b 18b

8 can be listed by equation (56a+16b) 8=18b

The solution is: 3a=2b, i.e., a b=2 3

Answer: The chemical formula of this oxide is Fe2O3.

3.A student weighed grams of potassium chlorate to produce oxygen, and in order to speed up the reaction, he also added a small amount of potassium permanganate, and when the reaction was completed, a total of grams of oxygen were prepared. Q: How many grams of potassium permanganate did the student add?

Solution: The equation for the decomposition of kclo3 to generate oxygen is as follows:

mno22kclo3*****==2kcl+3o2↑

then the mass that can generate O2 is (

That is, it is obtained by the decomposition of KMNO4.

From 2kmNO4*****K2MNO4+MNO2+O2:

The mass of potassium permanganate is (316

Answer: The quality of potassium permanganate is:

4.In a sufficient amount of hydrogen gas stream, after the mixture of hot copper oxide and elemental copper is completely reacted by 80 grams, the mass of the solid mixture becomes grams, and how many grams of copper oxide is found in the original solid mixture? And what is the mass fraction of copper in the original mixture?

Solution: According to the equation cuo+h2*****=cu+h2o, we can know that:

The mass of the reduced element is o.

So the mass of cuo in the original mixture is (

The mass fraction of the cu element is:

Answer: The mass of cuo in the original mixture is 32g;

The mass fraction of the cu element is 92%.

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2 The mass of water produced by the reaction containing oxygen (O) is.

Let the chemical formula of the substance be fexoy then 18y (56x+18y) =

If we get x y=3 4, then Fe3O4

kclo3==3o2+mncl+2kcl244 96

xx approx. = mass fraction of impurities =

Analysis: Because of the presence of sulfuric acid, sodium hydroxide was added without precipitation, and precipitation occurred after adding 25g, that is, the first 25g of sodium hydroxide solution reacted with sulfuric acid. After 25g of sodium hydroxide solution reacts with copper sulfate to produce precipitation, the mass of the precipitate is x, and the mass of sodium sulfate is y

2naoh + cuso4 = cu(oh)2↓ +na2so425g*16% x y

80/25g*16%=98/x=142/yx= y=

Let the mass of sodium sulfate produced by reaction with sulfuric acid be m

h2so4 + 2naoh = na2so4 + 2h2o25g*16% m

98/25g*16% =142/m

m = so the total mass of sodium sulfate produced =

Total mass of the solution after the reaction =

The mass fraction of the solute in the resulting solution after the reaction =

3) Add water. Let the mass of the added water be x, then: x=

Depending on the last substance generated is sodium sulfate in proportion to the amount of the substance2Na---SO4

50* xx=Let the amount of sulfuric acid in the mixed solution be m, and the amount of copper sulfate is n series of equations: m+n=

98m+160n=

The solution gives m= n=

1. The mass of the precipitated (copper hydroxide) grams.

2. The solute is sodium sulfate with a mass fraction of .

3. Add water to dilute the solution

A1. During the reaction of H2SO4+2NaOH=Na2SO4+2H2O, the mass of NaOH is: 25X16%=4G; Therefore, the mass of H2SO4 is calculated as: m1=4 80x98=; The mass of Na2SO4 is:

m2=4/80x142=

2. During the reaction of CuSO4+2NaOH=Cu(OH)2 +Na2SO4, the mass of NaOH is: (50-25)X16%=4g; The mass of copper sulphate is: m3 = 4 80x160 = 8g; The mass of Na2SO4 is:

m4=4/80x142=;The mass of Cu(Oh)2 produced by the reaction to precipitate is: m5=4 80x98=

3. The total mass of the solution in the solution after the reaction: Total solute mass in solution:; So the quality score is:

4. To make the mass fraction of the solute in the solution after the reaction 10%, it is necessary to add an appropriate amount of water, and the amount of water added is: (