Chemistry! Calculation and Analysis Questions, a chemical calculation question, asking for help with

Supplementary conditions: dilute hydrochloric acid is 73g dropwise to be suitable for the reaction completely) Because dilute hydrochloric acid contains 10% HCl, there is HCLN(HCl)=m(HCl) m(HCl)=Na2CO3 + 2HCl=2NaCl + H2O +CO2 x Y Z

So x= y= z=

Then the original mixture contains NaCl: =

Generate nacl:

Generate CO2: =

So the solution contains NaCl:

w% = =

Na2CO3 + 2HCl = 2NaCl + H2O + CO2 According to the above reaction formula, 73 grams of 10% dilute hydrochloric acid are consumed, and the hydrochloric acid consumed is grams.

The carbon dioxide produced is grams).

The resulting water is grams).

The solute before the reaction is grams).

The solute after the reaction is grams).

The solution after the reaction is grams).

The solute mass fraction in solution is.

1) The change in the valency of metallization can be judged by the electron conservation method.

n: no3- no2 , +5 +4, the valency decreases, and 6eo: no3- o2 , 2 0 , the valency increases, and the loss of 4e electrons in the redox reaction is conserved, so the metal element loses electrons and the valency increases.

3) The composition of the residual solid may be Cu2O or Cu2O, Cuo mixture (because the valency of Cu must be reduced, so there must be Cu2O) When the volume ratio of No2 and O2 generated by copper nitrate decomposition is 4:1, the valency of Cu remains unchanged, and according to the conditions in the question, it can be seen that the valency of Cu is reduced, so the volume ratio of Cu to must be less than 4:1, and when the mixed gas is dissolved in water, O2 must be excessive.

According to the conservation of n atoms, Cu(NO3)2 2Hno3 The concentration of the nitric acid solution (the resulting solution) = m 18 2 2m 9 mol l

4) To determine the composition of metal oxides, the extreme value method and the atomic conservation method can be used.

If all copper nitrate is decomposed into Cu2O, it can be obtained according to the conservation of copper atoms

cu(no3)2 ~ 1/2 cu2o

Then the mass of Cu2O = 188 2 144 = g If all the copper nitrate is decomposed into CuO, according to the conservation of copper atoms, it is obtained:

cu(no3)2 ~ cuo

Then the resulting mass of Cuo = 188 80 = g and the actual mass of the resulting oxide is, so the oxide should be a mixture of Cu2O and CuO.

Let the amount of substance of the mixture Cu2O be x and the amount of substance of Cuo be y, then 144x + 80y =

2x + y =

The solution yields x = mol and y = mol

Cu2O mass = * 144 = g

cuo mass = * 80 = g

mgso4+naoh=mg(oh)2↓+naso4，120 4025x 10%*w2

When the reaction is not complete, the amount of precipitation is proportional to the amount of sodium hydroxide added. When the equivalence point (set to w2 grams, which is not clear in the figure), the amount of precipitation generated does not change. Let the solute mass fraction of this solution be x.

The calculation formula is: x=

I can't see the numbers shown on the graph clearly.

20*80%=16t

caCO3===( high temperature) cao+CO2 (up arrow) 100 44

16t x100/16=44/x

x = A: The mass of carbon dioxide produced is tons.

The mass g of oxygen produced

Mass g of CO2 in the original gas mixture

Easy way to rush ideas.

The first question is the reaction that occurs.

2co2+2na2o2=2na2co3+o22h2o+2na2o2=4naoh+o2

Since we know that the gas mixture is absorbed, and the weight gain is only 6g, it must be caused by O2 running away, so the oxygen is g

The second question, 1 occurred reaction.

2co2+2na2o2=2na2co3+o22h2o+2na2o2=4naoh+o2

2 It would be easier for you to use the amount of matter, let CO2 be x mol, h2o y mol, and the mass of the gas mixture with h2o(g) is x * 44 + y * 18 = ;

According to the equation, the relationship between H2O and CO2 and O2 is 2:1, so the final oxygen is (x+y) 2 mol

Because the mass of oxygen g = 32 * x+y) 2 is solved to get x = mol i.e. CO2 is g

We can know that there are two reactions in this reaction:

2CO2+2Na2O2 2NaCO3+O2 2H2O+2Na2O2 4NaOH+O2Both can produce oxygen. At 120 °C, sodium hydroxide does not react with carbon dioxide to form sodium carbonate. So we have the following chemical equivalence:

2CO2 2NaCO3+O2 i.e. 1molCO2 will have 1molCO to become a solid, i.e. sodium carbonate.

2H2O 4NaOH+O2 i.e. 1molH2O will have 1molH2 to become a solid, that is, to become sodium hydroxide.

Then let the original carbon dioxide have xmol and water ymol. Then there is the following system of binary linear equations:

44x+18y=

38x+2y=6

The solution is x= y=

then oxygen mass =

Mass of carbon dioxide in the original gas mixture =

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Example 1There is a monochromatic light passing through a colored solution with a thickness of 1 cm, and its intensity is reduced by 20%. If the same solution with a thickness of 5 cm is passed, how much does it decrease in strength?

Solution: lgt1 c1 ——= ——lgt2 5c1

lgt2 = 5lgt1= 5× = -

t2= , i.e. weakened.

Absorbance is additive.

According to the proportional relationship of the reaction formula 2NO2=2NO+O2, it can be calculated that 1mol of NO2 must participate in the reaction of the oxygen generated.

Therefore, the initial concentration of NO2 is 1, and the decomposition rate is 1