A high school chemistry problem, seeking the process, a high school chemistry problem to process onl

Because LNA2CO3 reacts with BACl2 to form BAC3, and NA2SO4 reacts with BACl2 to form BASO4

BaCO3 reacts with dilute nitric acid, and Baso4 does not react with dilute nitric acid, so Baso4 is Na2CO3 is, is, and the concentration is.

The relative molecular mass of Baso4 is 233, so there is, and the concentration is 4mol L. The volume of the gas is.

Hope, thank you.

Na2CO3 + BaCl2 BaCO3 + 2NaClNa2SO4 + BACl2 BaSO4 +2NaCl Due to the excess of BaCl2, all carbonate and sulfate are precipitated, and the mass of the two precipitates is reacted with nitric acid.

baco3+h→h2o+co2↑+ba

So the reduced mass is the mass of baCO3, the quantity is, and the remaining mass of baso4 is, the amount is.

Therefore, it can be calculated that the concentration of Na2CO3 is.

The concentration of NaSO4 is:

So n(co2)=,v(co2)=n*

I feel that the question in the question should be "the concentration of two substances in the original mixed solution", if you can only answer the respective mass fractions of the two mixed solids according to your question, so I will answer the amount concentration of the substances in the two solutions.

Solution: (1) According to the meaning of the question, m(baso4) = , m(baCO3) = then n(baso4) = 233g mol=, n(baCO3) = Na2CO3 + BaCl2 == BaCO3 precipitation + 2NaCl1mol 1mol

n(na2co3)

1mol/n(na2co3) =1mol/n(na2co3) =

c(na2co3) =

Na2SO4 + BACL2 == BaSO4 (precipitate) + 2NaCl1mol 1mol

n(na2so4)

1mol/n(na2so4) =1mol/n(na2so4) =

c(na2so4) =

BaCO3+2Hno3==Ba(NO3)2+H2O+CO2 gas.

1mol 1mol

n(co2)

1mol/n(co2)=

v(co2)=n(co2)×vm=

1: The mass of Baso4 is obtained, the molar amount of Baso4 is 233g mol, the molar number of Baso4, the molar number of NaSO4, and the concentration:

BaCO3:,BaCO3 mol 197g mol,BaCO3 mol,Concentration:;

2: CO2 moles v=

This question is actually the flexible application of the chemical equation we have learned, I can't play this complete process, so I will write it for you briefly, if you don't understand it, please ask the equation mno2 + 4HCl (concentrated) = mnCl2 + Cl2 (gas) + 2H2O (this is water) The relative molecular mass of mnO2 is 87

So that's it, so the amount of matter of Cl2 should be not miscalculated, right? Finally, as for the following question, we can consider the calculation from the conservation of elements, and finally the Cl left in the solution is MnCl2 on the one hand, and NaCl on the other hand, so the amount of Cl is found from these two aspects, and the Cl in this is calculated according to the amount of NaOH) Yes, so all Cl together is. That is to say, the quantity of the substance of AGCL is, and the mass is.

That's it, if you need it, please ask

Principle: The law of conservation of mass and the law of conservation of charge.

This is a redox reaction, the valency of the reduced manganese element and the increased valency of the chlorine element are always equal, so the amount of chlorine gas generated is equal to the amount of manganese dioxide, which is mol (law of conservation of charge), and when reacting with silver nitrate, there are two forms of chloride ions in the solution: sodium chloride and manganese chloride, sodium chloride is calculated according to sodium hydroxide (mol), manganese chloride is calculated according to manganese element, there is mol. There are a total of mol of chloride ions in the solution, so the resulting silver chloride also has mol, and finally it is changed to mass...

1.According to the chemical equation 1molmno2 1molCl2, is, Cl2 is.

1mol LNaOH has, the remaining hydrochloric acid is also there, and MnCl2 is, AGCL precipitate can be generated.

(1) Discoloration of the solution; c2h4+br2=(ch2br)2;nch2=ch2---ch2-ch2]n

2）；100%；Two.

Process: Atomic utilization (molecular weight of the expected product and sum of the molecular weight of all products) 100

Process 1 44 (44 + 111 + 18) =

Process two 44 44 = 100%.

(1) Fading, color fading Equation Addition Addition Equation.

The color of the carbon tetrachloride solution of bromine is discolored C2H4+BR2=(CH2BR)2 NCH2=CH2---CH2-CH2]N

100% choose the second option.

If the two equations of 2NO+O2=2NO2 and 3NO2+H2O=2HNO3+NOno are calculated, there is a surplus of no, and the result may be 46% (not calculated).

But with the combined equation of these two equations, 4NO+3O2=4HNO3, NO is all transformed.

Suppose 1mol of ammonia --- produce 1mol of NO--- 1mol of nitric acid is generated, and 1mol of ammonia is required to neutralize.

Question 1 should be 50% upstairs

n nh3=100000/17mol

According to the above calculation, the ammonium nitrate generated should be 100000 34molm NHNO3=100000 34*80 g=