Chemical calculations with equations and processes .

Reaction: Na2SO3 + H2SO4 = Na2SO4 + SO2 + H2O [definitely].

Na2SO4 + BACL2 = BaSO4 +2NaCl [definitely].

Na2SO3 + BACL2 = BaSO3 + 2NaCl [possibly, when sodium sulfite is in excess].

H2SO4 + BACL2 = BASO4 + 2HCl [possibly, when there is an excess of sulfuric acid].

Among them, the above reaction needs to discuss the excess of Na2SO3 or the excess of sulfuric acid on a case-by-case basis, so two reactions are introduced.

The amount of sulfuric acid substance n=cv=

Let the amount of the original Na2SO4 substance x, and the amount of Na2SO3 substance Y

Then the amount of SO4 2- in the original solution is x, and the amount of SO3 2- is y

When sulfuric acid is added, SO42- is added, and there is no actual reaction.

Therefore there must be and only (x+SO4 2- is finally precipitated.

First of all, the first equation can be listed, fixed, according to n = m m, with the mass of the initial solid mixture as the equivalent:

1）142x+126y=

a) If there is an excess of Na2SO3, then H2SO4 is depleted, and the amount of Na2SO3 species that is reacted away and the amount of remaining SO3 2- species (

This is also the amount of Baso3 substance.

Then according to n=m m, with the precipitated mass as the equal quantity, the column equation (2).

2）233(x+

If it is solved in conjunction with (1), it can be solved to x 0, which does not conform to the objective facts, so this situation is not established.

2) If there is an excess of H2SO4, then Na2SO3 is depleted; That is, there is only baso4 in the pellet

Hence according to n=m m

233(x+, solution x=; There is no need to solve y, because the problem is called Na2SO4, and according to n=m m, the mass fraction is.

The following question is exactly the same ...... as yours

Na2SO4=> then the amount of matter is.

Then the amount of Na+ substance is.

Since both contain two Na+, the amount of the common substance is , let Na2SO4 be Amol, then: (46 + 96) A + (46 + 80) (the solution can be done.

The answer in the question is incorrect.

Only HCl, which produces chlorine Cl2, is oxidized. The formation of KCl and CrCl3 is not considered to be oxidized.

As can be seen from the chemical reaction equation, only 6 of the 14 chloride ions are oxidized to chlorine atoms.

So it's calculated like this.

The parentheses (14*) are changed to (substitute data, m2=

1.The lab has a 500ml bottle of waste containing sulfuric acid, which happens to neutralize this sulfuric acid and consumes sodium hydroxide. What is the amount of H2SO4 in the waste liquid?

2naoh + h2so4→ na2so4+2h2o

The amount of H2SO4 in the xmol waste liquid x=

2.Hydrogen can be produced by reacting 100g of hydrochloric acid with a sufficient amount of iron. What is the amount of HCI in this hydrochloric acid solution?

fe + 2hci→feci2+h2↑

The amount of HCI substances contained in xmol hydrochloric acid solution x =

3.The students of the Mona Noise Technology Group mixed 16g of copper oxide with excess carbon powder evenly, what is the maximum amount of copper that can be obtained in the experiment?

C+ 2CuO (high temperature) 2Cu + CO2

2*80 2mol

16g xmol

The maximum amount of material that can obtain copper x =

4.In the Mo solution samples taken from the electrolyzer of the Mozambican chemical plant, the solutes were mainly sodium hydroxide and sodium chloride (the other solutes were negligible), which happened to be completely neutralized when H2SO4 was added. How many grams of sodium hydroxide is in a 25ml sample of this solution?

2naoh + h2so4→na2so4+2h2o

2*40 1mol

xgx = 3g25ml How many grams of sodium hydroxide are in the sample of this solution?

ps: , uh, I don't know too much where your 25ml sudden-jin eggplant came from.

For every planned production of 30,000t of iron (including Fe97%), 1) if hematite containing Fe2O3 70% is used to produce, how many tons is required?

2) How much t is needed to switch to magnetic ore containing 70% Fe3O4 for production?

You can find the answer in the workbook.

Assuming that the mass of CaCO3 in each tablet is X, then.

caco3+2hcl=cacl2+h2o+co2100g 44g

x100g : 44g = x :

x = where the mass of the ca element is: x 40 100 = , i.e. 500mg

Solving 9 grams of water can produce x grams of hydrogen from celery.

2h2o=2h2+o2

xx=(4×9)/36=1g

Solution: The amount of the first hole of the mass grip that produces O2 is x grams.

2h2o=2h2o2

xx=32×

To dissolve, the section needs to be decomposed x grams.

Hydrogen peroxide. 2h2o2＝2h2o+o2

xx=160×68÷32＝340g

1,2h2o==2h2+o2

9g...1g..8g

2.The quality of the hydrogen nucleus lead gas or cherry blossoms: change the group well.

2h2o==2h2+o2

.2h2o2==2h2o+o2

340g...160g