To the process, the process is good and rewarded with 50 points

The total precipitated grams, after adding nitric acid, reduce the grams to obtain grams of barium carbonate (BaCO3), which should be Baso4 insoluble in any substance! BAC3+2Hno3=BA(NO3)2+H2O+CO2 "The amount of some substances below carbonic acid is 187, write grams below", "write 44 below carbon dioxide and write x below" and then multiply diagonally to equal, find the mass of carbon dioxide, and require the volume of gas v=(x 44)*! Then find the concentration of sodium carbonate!

The amount of the substance obtained is the amount of the substance of sodium carbonate, and then divide by 50 to get the concentration! Then find the concentration of sodium sulfate with (the amount of barium sulfate substance, that is, the amount of sodium sulfate, the amount of the substance, and then divide by 50 to get the sodium sulfate concentration!)

Solution: LZ is good!

First of all, the first line is na2co3

Started, the important condition of the second line is that "the precipitation is reduced to indicate that these precipitation are baso4."

nbaso4=。Then write out the reaction:

Na2SO4+BACl2==2NaCl+BaSO4 so N1=.

Then the reduced precipitate (all BACo3. That is, there is.

nbaco3=, write the reaction:

Na2CO3+BACl2==2NaCl+BaCO3 so N2=.

That is, the amount and concentration of Na2CO3 is;

Na2SO4.

Question 2: There is basically no problem now Write out the reaction, and the data will be basically done.

BaCO3+2Hno3==Ba(NO3)2+H2O+CO2 So N3=.

So the release of CO2 is, VCO2=.

So the volume of the resulting gas in the standard condition is pull

Na2CO3 + BACl2=BaCO3 (precipitation) + 2NaClNa2SO4 + BACl2=BaSO4 (precipitation) + 2NaClBAC3+2HNO3=BA(NO3)2+CO2 (gas) + H2O Take 50ml of mixed solution of Na2CO2 and Na2SO4, add excess Baci2 solution to obtain grams of white precipitate, and the white precipitate is a mixture of BaCO3 and Baso4.

After treatment with excess dilute nitric acid, the precipitate is reduced to grams and gas is released.

The remaining is Baso4, so the mass of BaCO3 is , and thus CO2 is obtained.

The original solution contains Na2CO3 Na2SO4

1.[na2co3]=

na2so4]=

2.The volume of gas under STP is:

The cosine of the angle of the vector is equal to the product of the number of vectors divided by the product of the modulus, and since it is an acute angle, its value is greater than zero, which can be calculated.