Eager to solve two junior high school chemistry problems!!

1. Analysis: Putting quicklime into water will react and produce mature lime, so it can be determined that the solute in its solution is Ca(0h)2:

cao + h2o == ca(0h)2

Let quicklime react with water in xg and form yg hydrated lime, cao + h2o == ca(0h)2

x y56 / = 18 / x

The solution yields x=56 = 74 y

The solution of y = according to the meaning of the question, it can be seen that Ca(0h)2 is not completely soluble in (water), therefore, the mass fraction of the solution at 33 degrees Celsius can be directly calculated by its solubility.

Solution: At 33 degrees Celsius, the mass fraction of the solution is:

For an exact answer, see the one above.

O2 is 32g, that is, 1mol, H2 and CO are 62g-32g=30g, if the oxygen reacted with H2 is xmol, then the reaction with CO is (1-x)mol2H2+O2=2H2O

2x x 2co + o2 = 2co2

2(1-x) (1-x)

2x*2+2(1-x)*28=62-32

x=1 2 so h2 has 1mol 2g, co is also 1mol 28g, oxygen 1mol 32g

Spoiled NaOH contains NaOH and Na2CO3, and it is clear that NaOH is not involved in the above reactions.

So the amount of NaOH that has deteriorated is obtained by Na2CO3 back-calculation, and the meaning of the title is to find the mass of NaOH in the mixture, that is, the unspoiled NaOH is used.

Got it As for the PH problem, there are still Naoh, NaCl and CaCl2 in the cup Junior high school does not consider the hydrolysis of salt [except for Na2CO3.

So ph>7

Sodium hydroxide does not react with calcium chloride.

The resulting precipitate is calcium carbonate produced by the reaction of sodium carbonate and calcium chloride by the deterioration of sodium hydroxide.

na2co3+cacl2=caco3+2naclx 5gx=

So the mass of NaOH in the sample is.

2) The pH 7 of the solution in the beaker is exactly reflected when the NaOH value after the reaction, and the mass of NaOH in the sample is the mass of sodium hydroxide that has not deteriorated. Not the quality of sodium hydroxide that has deteriorated.

(1) First of all, NaOH has deteriorated into Na2CO3, so there is the following reaction:

Na2CO3 + CaCl2 = CaCO3 + 2NaCl From this, the mass of Na2CO3 is , and the mass of CaCl2 participating in the reaction is .

Therefore, the NaOH mass is:

2) It happens to be completely reacted, at this time the solution is left with Ca2+ and Cl-,Ca2- hydrolysis, so the pH <7.

is to find the value of NAOH in the sample before the reaction.

(1) First of all, NaOH has deteriorated into Na2CO3, so there are the following reflections:

Na2CO3 + CaCl2 = CaCO3 (precipitate) + 2NaCl From this, the mass of Na2CO3 can be obtained, and the mass of CaCl2 participating in the reaction is .

Therefore, the NaOH mass is:

2) It happens to be completely reacted, and the solution is left with Ca2+ and Cl-, so the pH < 7.

is to find the value of NAOH in the sample before the reaction.

When carbon dioxide is generated, the mass ratio of carbon to copper oxide is 12:160, when carbon monoxide is generated, the mass ratio of carbon to copper oxide is 12:80, and the mass ratio of carbon to copper oxide in the question is 12:100, so C should be selected.

Yes, first it is obvious that copper oxide is in small amounts, so a second reaction will occur to generate grams of copper with a remainder of 2 grams of copper.

The CO and CU reactions will produce carbon dioxide, but the remaining CO will be yours to do the math.

You can't be saved, you won't be able to answer such a simple topic, it's just an overdose of toner, and you want to go!

Because copper oxide is and carbon is copper oxide, if 1 is carried out, cuo is insufficient, and if 2 is carried out, c is insufficient, so both gases are present.