Science Problems Chemical Calculations for junior high school

Explain your thinking, it's not that you don't react.

Analysis: Because Cu originally generated a patina, this copper is a mixture of Cu2(OH)2CO3 and elemental Cu, right, when heated, the patina decomposes, and the solid mass decreases, but the mass of Cu and O2 reacts to Cuo.

Now it's equal, so that's the key to this calculation.

Basic copper carbonate is heated to produce black solid copper oxide, water vapor and carbon dioxide, in which water vapor and carbon dioxide volatilize and the mass is reduced. However, copper is heated in the air to form copper oxide, and the mass increases. Both the reduced mass of water vapor and carbon dioxide is equal to the mass of oxygen that reacts with copper.

Let's make this equality an unknown y g

Let the mass of the original mixture be 100g, of which the basic copper carbonate is xg, then the copper is 100 xg, according to Cu2(OH)2CO3 2CuO+H2O+CO2 xg YG222:(18+44)=x:y

y=62x/222

2cu + o2 ===2cuo

100-xg yg

128/32＝（100－x)/y

y=32*(100-x)/128

The y of the above two equations is equal so.

62x/222=32*(100-x)/128x=

The mass fraction of Cu2(OH)2CO3 in the original mixed powder is.

2cu + O2 = Heating = 2Cuo Solid mass increases.

Cu2(OH)2CO3 = heating = 2CuO+H2O+CO2 Solid mass decreases.

x 32222/x=62/32

x = mass fraction of basic copper carbonate in the original mixed powder.

Basic copper carbonate is thermally decomposed to form copper oxide, and the mass decreases.

Copper is heated and reacts with oxygen in the air to form copper oxide, which increases in mass.

Let the copper oxide participating in the reaction be xg. The resulting copper is 64 (x 80) g and the unreacted copper oxide is (10-x) g

Equation: 64(x 80)+(10-x)=

solution, x=8Therefore, the amount of copper oxide participating in the reaction is 8 grams.

After heating, the solid is a mixture of copper oxide + copper, then the mass difference between the front and back is the lost o atomic mass:

o atom i.e. o atom i.e. cuo is involved in the reaction, i.e., 8g cuo

H2 + Cuo = Heating = Cu + H2O Solid Mass Reduction 80 64 80-64 = 16

x80/x=16/

x=8g

Solution: The solute potassium carbonate in the potassium carbonate solution with a solute mass fraction of 50% is: grams, solvent water grams;

Let the calcium carbonate be generated x grams, and the potassium chloride Y grams will be generated, according to the reflection CaCl2 + K2CO3 = CaCO3 + 2KCl

x y138 ：100= ：x 138 ：149 = ： y

The solution yields x = 10 grams, y = grams.

That is, the mass of calcium carbonate precipitate is: 10 grams, and the total mass of the solvent (water) in the final solution is: grams, according to the solubility at t temperature, 40 grams of water dissolvable potassium chloride is: grams, and the reaction generation grams are greater than the soluble grams, so there will be potassium chloride crystals precipitated in the solution, and the final solution obtained is a saturated solution at this temperature, and its concentration is: Description: This question mainly needs to consider how much potassium chloride solute can be dissolved by water in the solution at this temperature, and whether there will be crystal precipitation, This requires judging the total mass of potassium chloride in the entire process. This is also the main purpose of the question given to the condition "solubility of potassium chloride at t degrees Celsius."

First of all, the potassium carbonate content in the potassium carbonate solution added grams, because calcium chloride + potassium carbonate = calcium carbonate (precipitation) + potassium chloride use this to calculate the quality of calcium chloride, calcium carbonate, potassium chloride, and finally consider the solubility Whether it can be completely dissolved Whether there is excessive potassium chloride I don't have an element quality table here to calculate it specifically The attention points are all there, and I won't do it, just ask me

Solution: Let the mass of the solute in the sodium hydroxide solution be x

hcl + naoh == nacl + h2o40

73g*10% x

The solution yields x=8g

8g/ 40 g/mol=

8/80g=10%

Answer: The amount of solute in the original sodium hydroxide solution is, the solute mass is 8g;

The mass fraction of the original sodium hydroxide solution is 10%.

Isn't this question fully described? For example, what is the use of "after evaporating the water from the reaction solution"?

hcl + naoh=nacl+h2o40

73g*10% x

x = 8g of the amount of solute in the solution of primary sodium hydroxide. 8g (40g mol) = mass fraction of the original sodium hydroxide solution. 8g 80g*100%=10% After evaporating the water from the reaction solution, when it is cooled to 20°, the solution happens to be saturated.

This question is useless.

naoh+hcl=nacl+h2o

40x 73*10%, x = 8 grams.

The amount of solute in the solution of primary sodium hydroxide = 8 grams.

Mass fraction of original sodium hydroxide solution = 8 80 = 10%.

mgcl2+2koh=mg(oh)2+2kcl2；Knowing the mass disturbance return and total mass of the filtrate, the mass of Mg(OH)2 after the reaction was found to be (50+; By using the reaction formula, the mass of koh can be found to be 2*m(koh)*; It is known that the mass of the solute in the solution is 25%, and the mass of the solute (mixture of sodium hydroxide and potassium chloride) is found to be 50*25%=; then the mass of potassium chloride is.

Solution: Let the mass of the solute in the KOH solution be X, and the resulting KCl is ym (solution) = 50g+

2 koh + mgcl₂ =2 kcl + mg（oh）₂↓x y

112 ： 58=x ：：y

x= y=The mass of the solute in the solution after the reaction is w=pai=

So it turns out that the solid mixture kcl - = g

20 g Answer: ....Spring Dahe .........Imitation ......

The product is magnesium hydroxide, so the mass of potassium hydroxide can be calculated, and thus the mass of potassium chloride.

Analysis: There is precipitation formation in this reaction, so the mass of the solution obtained after the reaction is equal to: the mass of the sodium hydroxide solution participating in the reaction + the mass of the copper sulfate solution - the mass of the precipitate.

The mass of the copper sulfate solution and the mass of Cu(Oh)2 and Na2SO4 can be calculated respectively.

Solution: Let 100g of sodium hydroxide solution with a solute mass fraction of 8%, add x grams of copper sulfate solution with a mass fraction of 20%, and the mass of the precipitate generated by the complete reaction is y grams, and the mass of copper sulfate generated is z grams.

cuso4+2naoh=cu(oh)2↓+na2so4

20% x 100*8% y g z g.

160/（20%x ）=80/（100*8%） =98/y=142/z

x=80y=z=According to the law of conservation of mass of chemical reactions, the mass of the solution obtained after the reaction is:

100+ grams.

The mass fraction of the solution obtained after the reaction is:

Answer: The mass of the precipitate generated by the reaction is grams, and the mass fraction of the solute in the solution obtained after the reaction is.

Precipitation Mass:

Copper sulphate quality:

Original copper sulfate solution quality:

Sodium sulfate mass:

Soluble mass fraction:

precipitation, and the mass fraction of the solution obtained after the reaction is.

As long as you calculate the mass of sodium hydroxide, find its molar amount, and calculate it according to the molecular weight and proportion of each substance in the equation, use your brain.

1 m total = 40 g+

m CO2 = m total.

2 Let the quality of caco3 be x for participating in the backhand

caco3+2hcl=cacl2+h2o+co2x

Solution x = 5g

5g 4 = Answer: The content of calcium carbonate in calcium tablets is true.

1) Because the solution is neutral and antecedent, it means that it should be completely reversed.

2NaOH + H2SO4 = 2 H2O + Na2SO415%*40G xg The solution yields x=

2) mg + 2HCl = mgCl2 + H2 xg x=

Mgo + 2HCl = MGCL2 + H2O This reaction does not produce hydrogen.

Hence the mass fraction of elemental magnesium in the magnesium zone.

Ecological spring answers into the quality of carbon dioxide.

caco3+2hcl=cacl2+h2o+co2x

x=5g5g/4=>

1) Resistant to jujube chaos m(CO2) = 40g + 4 * rock transport set the mass of CaCO3 is x, then 100 x = 44 x = 5g a piece of calcium tablets The mass of a piece of calcium tablets is 5g 4 = The content of calcium in a piece of calcium tablets is 40 100 * 100% * * Therefore, it is a false Chang file.

106 g of the filtrate is Na2CO3 solution.

According to the law of conservation of mass, the mass of CO2 gas produced by the reaction = 106+ =2HCl + Na2CO3 ==2NaCl + H2O + CO2 x Y Z

73 x = 106 y = 117 z = 44 x = , y = , z =

1) The mass fraction of the solute in the resulting gram solution = =2) The mass fraction of the solute in the added hydrochloric acid = =

3) From the above calculation, it can be obtained that the mass of Na2CO3 in 106g of Na2CO3 solution is so the mass of Na2CO3 in Na2CO3 solution: the mass of water = :( = 1:4, and because gram water is added at the beginning.

So the actual mass of Na2CO3 in the 50g sample ==, so the mass fraction of sodium carbonate in the sample ==

Question question, right? I don't know the concentration of hydrochloric acid, how to calculate it!

2LiOH (solid) + CO2 (gas) = Li2CO3 (solid) + H2O (liquid) 48 44

1g x x = (1) Try to calculate the mass of CO2 that can be absorbed by 1g of lithium hydroxide 2KOH (solid) + CO2 (gas) = K2CO3 (solid) + H2O (liquid) 112 44

1g y y = (2) the amount of CO2 that can be absorbed by 1g of potassium hydroxide.

3) Lithium hydroxide of equal mass absorbs more CO2 than potassium hydroxide, so the spacecraft uses lithium hydroxide to absorb CO2

4) The 4 astronauts released a total of CO2:502, i.e., 2LiOH (solid) + CO2 (gas) = Li2CO3 (solid) + H2O (liquid) 48 44

z z=

na2co3＋cacl2 === caco3↓＋2nacl106 111

ag bg×10%

a/b=na2co3＋2hcl2 === nacl＋h2o＋co2↑106 73

ag m(hcl)×10%

m(hcl)=73a/

m(hcl)＝

Therefore, the mass of hydrochloric acid is twice that of BG.

2LiOH (solid) + CO2 (gas) = Li2CO3 (solid) + H2O (liquid) 2*24 44 74 18 (1) 1g lithium hydroxide absorbs CO2: 44 (2*24) = (2) In the same way, 1gkoh absorbs CO2: 44 (2*(3) The less mass of matter carried on the spacecraft, the better, obviously, Lioh is far better than KOH (4) 4*502*7