Chemistry problems in the first year of junior high school calculation problems solution mass calcul

1) m=3 The reason is: after adding 10 grams of water for the second time, there is potassium chloride that is not dissolved (because it is dissolved after adding water later), comparing the remaining solid mass of 12 two times, it can be seen that every 10 grams of water can be dissolved, and the fourth time leaving 3 grams of insoluble, which is only different from the second time, indicating that 3 grams are all mno2, so the remaining mass of the third addition of water is 3 grams.

2) 3) Solution: Oxygen xg is set

2kclo3=mno2=2kcl + 3o2 (and a heating symbol) x 245 x=

Answer: Oxygen is generated.

(1) The dilute hydrochloric acid taken by B and the sample happen to be completely reacted, and the mass of the beaker and the dilute hydrochloric acid plus the mass of the sodium chloride added minus the mass of the beaker and the remaining material after the reaction can be used to know the mass of carbon dioxide generated. Such as:

75g+, 65g+, 65g+, it can be seen that the mass of carbon dioxide generated is that carbon dioxide is generated when the sample is 4 grams, indicating that 75g of dilute hydrochloric acid must be more, and 65g of dilute hydrochloric acid may be just the end of the reaction; In the second and third cases, there is the same amount of hydrochloric acid, the sample is 4g, and 5g both generate carbon dioxide, indicating that the sample is 5g and the sample is 4g, which is exactly the complete reaction. In summary, the hydrochloric acid taken by student B happened to react completely with the sample.

2）na2co3---co2

x106 x=44 So x=, which is the mass of sodium carbonate in the sample, is.

Then the mass of sodium chloride is and the mass fraction of sodium chloride in the sample is.

3)na2co3+2hcl=2nacl+h2o+co2

y117/y=44/ y=

The total mass of NaCl in the beaker after the reaction is that because the sample reacts fully with dilute hydrochloric acid, only NaCl and water are left in the beaker after the reaction, so the mass fraction of the solute in the solution obtained after the reaction is.

Student B is 65+

2hcl+na2co3===2nacl+ co2+ h2o

x= y=sodium chloride mass fraction.

Solution solute mass fraction: (

Analysis: 1. By comparing A and B, it is known that there is an excess of hydrochloric acid in A, and then comparing B and C, it is known that there is an excess of samples in C, so B's reaction happens to be complete.

2. According to the law of conservation of mass, the carbon dioxide produced by Zhiyi is: 65g + 4g - so as to find the mass fraction of sodium chloride in the sample.

3. There are two parts of sodium chloride** after the reaction, one is in the sample, and the other is generated.

Solution: (1) The dilute hydrochloric acid taken by student B happened to react completely with the sample.

2) Let the mass of sodium carbonate in sample B be y1

2hcl + na2co3 === 2nacl + co2↑ +h2o

y1 65g + 4g --

Find y1===

So the mass fraction of sodium chloride in the sample is:

4g --4g x 100%

= (3) Let the mass of sodium chloride generated in B be y2

2hcl + na2co3 === 2nacl + co2↑ +h2o

y2 65g + 4g --

Find y2===

So the total mass of sodium chloride in solution is:

4g --= According to the law of conservation of mass, the mass of the solution is:

- 25g ===

So the mass fraction of the solute obtained by the solution exactly after the complete reaction is:

x 100%===

Student B is correct and completely reacts, using the data of the first two rows and subtracting the third row as the CO2 mass, the column equation is solved.

Carbon dioxide is generated.

na2co3+2hcl=2nacl+h2o+co2x y

106/x=117/y=44/

x= y=1, the original mixture NaCl

2. The mass fraction of the solution after the reaction. (

If you make a mistake, the mass of the solution obtained is greater than the sum of the original three masses. Wrong, don't do it.

Saturation, potassium nitrate excess, solution 50 + 15 = 65 g

Mass fraction 15 65 * 100% =

Suppose the mass of the diluted solution is xg, where the mass of the solute remains unchanged, all have x*, obtain, x=3500g, so the mass of water added = 3500-1000=2500g

1kg of concentrated sulfuric acid with a solute mass fraction of 98%, of which H2SO4 has a mass of 1 98%=

Diluted into dilute sulfuric acid with a solute mass fraction of 28%, the solution mass is 980 28% = 3500 grams.

So the mass that needs to be added is 3500-1000=2500 grams.

The mass of sulfuric acid in the original solution is 1000*98%=980g, and the quality of the water in the original solution is 1000-980=20g

The total mass of the prepared solution is 980 28% = 3500g, and the mass of water required is 3500-20 = 3480g

Because, at 20 °C, the solubility of sodium chloride is 36 grams, that is, 100 grams of water can only dissolve 36 grams of sodium chloride at most, and now 40 grams of sodium chloride are put into 100 grams of water, it can not be completely dissolved, and the mass of the solution can only be obtained: 100 + 36 = 136 grams.

Similarly, 36 grams of sodium chloride only requires a mass of 100 grams of water to be completely dissolved.