[Thematic objective]
1。 Master the use methods and skills in chemical calculation.
2。 Strengthen basic calculation skills, improve ability, and use fast calculation and clever solution and mathematical calculation methods. [classic question]
Two problems: different application methods
[Example:1]10ml gaseous hydrocarbon used for complete combustion is restored to its original state (gas volume measured by 80 ml oxygen1.0105pa at 27℃) of 70 ml, and the molecular formula of the hydrocarbon is obtained.
Coach, after the reaction, the total volume of 90 ml raw gas mixture is 70 ml, and the volume is reduced by 20 ml. The residual gas carbon dioxide and excess oxygen that should be produced can be calculated by the following difference method.
CXHY +(X+) reduces the volume of O2 xCO2+H2O.
1 1 +
10 20
Calculate the molecular formula of hydrocarbons with y = 4/ > C3H4, C2H4 or CH4[ legal abstract]
The difference method is to determine the so-called "theoretical lean substance" according to the chemical equation or relationship before and after a certain amount of change, which can be the mass difference, the volume difference of gaseous substances, the pressure difference, the amount of possible substances, the heat and lean in the reaction process. This method is suitable for the answers in the reaction mixture, the answers in the reaction system and the differences before and after the reaction.
[consolidate the use
1, the existing mixture of KCl and 3.87 g KBr was completely dissolved in water, and 6.63 g of precipitate was produced after adding excessive silver nitrate solution for full reaction. The mass fraction of potassium in the original mixture was
a 0.24 1 b 0.259 c 0.403d 0.487
Problem: Law of Conservation
Example 2 The same mixture of copper, cuprous oxide and copper oxide 100 ml was added to the solution for dissolution, and 224 ml of NO gas was collected (under standard conditions). Looking for:
(1) Write the ionic equation of the reaction between cuprous oxide and dilute nitric acid.
(2) The amount of finished copper nitrate.
(3) If the mixture contains 0.0 1moLCu, it is characterized by obvious amounts of cuprous oxide and cupric oxide?
(4) Calculate the amount of cuprous oxide and the range of X for the mixture, such as the amount of copper. ..
[Coach] The topic is "Mixed Computing". If the equation holds, the process of solving the problem is more complicated. Grasping the initial state and final state of the reaction can solve the relationship between protection and achieve the purpose of simplification.
(1) Use electronic conservative decoration. 3cu 2 o+ 14 HNO 3 = = 6Cu(NO3)2+2NO ↑+ 7 H2O
(2) protection N(HNO3)== 0.06 mol, N(NO)== 0.0 1 mol,
N (Cu(NO3)2)==(0.06 N atom. -0.0 1)/ 2 = 0.025
(3) The title mixture contains three substances, copper, cuprous oxide and CuO, but only participates in the redox reaction of copper and cuprous oxide, so we use electron conservation to solve it directly.
Electron transfer of total number: n (e-) = n (no) × 3 = = 0.03mol.
The number of electrons in copper: 0.0 1×2 = 0.02 mol.
The number of electrons of cuprous oxide: 0.03-0.02 = 0.0 1 mol n (cuprous oxide) = 0.0 1/2 = 0.005mol.
N (CuO) = 0.0025-0.01-0.005× 2 = 0.005 mole.
(4) According to the solution of (3), electronic protection is provided based on the assumption that n (cuprous oxide) = 0.0 15-Xmol N(CuO) is the extreme value of X-0.005mol: when the electron changes from copper to n (copper) = 0.015 mol; Electronic cuprous oxide Cu2O has n () = 0.0 15mol, positive (Cu2+) == 0.03mol greater than 0.025mol, positive (Cu) is not equal to 0.005mol, others = X- pair n(CuO) is greater than 0 and positive (Cu) is 0.005mol..0.005
[Method Summary]
The recombination of atoms in chemical reactions has a series of protective phenomena: the law of conservation of mass, the conservation of elements, the conservation of charge, the conservation of chemical reactions and the conservation of electron gains and losses according to the law of conservation of mass. Therefore, the solutions of these conservation laws are called conservation laws. The total mass of substances with conservation of mass has the same chemical reaction before and after, and the solution prepared or diluted in this process has the mass of solute constant. The types of elements with conservative reaction constants before and after have the same atomic number, the same amount of substances and the same mass. The charge of any one of them is conserved, that is, an electrically neutral system, such as a compound, a solution in a mixture, etc. Algebraic zero of charge, that is, the sum of positive and negative charges is equal to the conserved total. The advantages and disadvantages of electrons are the occurrence of redox reaction. The number of electrons obtained by oxidant must be equal to the number of electrons obtained by reductant, and the spontaneous redox reaction or primary cell or electrolytic cell is lost.
Integration]
Copper carbonate, basic copper carbonate, can be dissolved in hydrochloric acid to form copper chloride, which can be decomposed into a mixture of copper oxide at high temperature. Dissolving 28.4g consumes 500ml 1mol/L hydrochloric acid, and the combustion quality of the mixture and the quality of copper oxide are compared.
a . 35g b . 30 c . 20d 15g。
Q: Legal relationship.
In order to prevent iodine deficiency disorders, the state stipulates that each kilogram of salt should contain 40 to 50 mg of potassium iodate. In order to test whether a certain salt is qualified iodized salt, a classmate took 428 grams of salt sample and tried to dissolve all potassium iodate. Acidify the solution, add enough potassium iodide starch solution to make the solution blue, and then titrate with 0.030mol/L sodium thiosulfate solution, leaving 18.00mL of blue that just disappeared. Test and calculate qualified iodized salt products. But the reaction is as follows:
IO3-+5 I-+6 H +→3I2 +3 H2O I2 +2
Thiosulfate ion ON-→2I-+ S4O62-
[Coach] The topic is "Multi-step reaction of calculation". According to the reaction equation, IO3- and S2O32 are directly established, which are related to the solution.
Solution: 6S2O32-IO3-
6 mol 1 mol
×× 10-3 of 0.030 mol/L.
18.00 ml n (I2) = = 0.09× 10-3 mol/>; KIO3 per kilogram of salt:
Qualified iodized salt
[Method Summary]
The actual chemical production and chemical research often involve a multi-step reaction: the reaction goes through several steps, and the content determination of a substance from raw materials to finished products can go through several intermediate steps. For multi-step reaction system, the relationship between the amount of starting material and final material is determined according to several chemical reaction equations, and the calculation method of column ratio is called "relational" method. Relationship method can save unnecessary intermediate calculation steps, avoid miscalculation and get the results quickly and accurately. The key to solving the problem is the relationship to be established: 1, the particle protection relationship, the equation addition of the number relationship of the stoichiometric formula (3) established by the relationship, and the relationship established by the relationship.
consolidate
As we all know, the bones of vertebrates contain phosphorus. The following is an experimental method to determine the phosphorus content in animal ashes. Take 0. 103 g sample animal ashes, treat them with nitric acid, then convert phosphorus into phosphate ester, and then add them into phosphate precipitation reagent. After 0.504 g of P2Mo24O77, the solid composed of precipitate (molecular weight: 3.60× 103) was ignited. The test mass fraction of phosphorus in ash was calculated from the sample data (the relative atomic mass of phosphorus was 3 1.0). )
Two problems: average method application program
[Example 4] Two of the four metals of zinc, iron, aluminum and magnesium are mixed with 10g and enough hydrochloric acid to produce hydrogen gas of11.2l. Under standard conditions, the mixture must contain metal ions.
Iron, zinc, aluminum, magnesium,
[Coach] Under the standard condition that electrons and the average molar mass of 10g metAl react with enough hydrochloric acid to generate hydrogen, the average molar mass of electrons of metal can be10g/mol1.2 liters, which are: zinc: 32.5g/mol al: 9g/mol mg.
Summarize regularly
The so-called average method is a mathematical method to solve problems in the principle of chemical calculation. It is an arithmetic average based on mathematical principles. These two numbers, MR 1 and MR2(MR 1 greater than MR2), must be somewhere in between. Mr. Average Sum has determined the value range of MR 1 and MR2, and the correct answer can be found quickly in combination with the topic. Common methods include: finding the average formula of average atomic weight, average molar electron mass and average composition.
[integration]
4.CO20. 1mol is generated by the reaction of 10g calcium carbonate containing impurities and sufficient hydrochloric acid. This example may contain impurities/>; A. ammonium bicarbonate and magnesium carbonate B. magnesium carbonate and silicon dioxide
C.K2CO3 and SiO2 D. cannot be calculated.
Various problems: the law of extreme assumptions
[Example 5] In a certain mass of magnesium, zinc and aluminum mixture, there is enough dilute H2SO4 to generate H2 2.8 L (under standard conditions), and the mass of the original mixture may be
a . 2g b . 4g c . 8g d . 10g br/>； [Coach] This title is due to lack of data. It can't find the mass of each metal, only within a certain range. Among the three metallic zinc, the aluminum material with the lowest mass has the same amount of hydrogen. Suppose the holiday of metallic zinc is 8. 125g, and the actual mass of all assumed metallic aluminum is between 2.25g and 8. 125g ... so the answers are B and C.
Summary of law]
"Extreme Hypothesis" is a commonly used mathematical method to solve chemical problems. The general answer to the calculation of mixture can be assumed to be a mixture of pure substances, and the determined maximum and minimum values can be calculated, and then analyzed, discussed and summarized.
consolidate
5. 0.03 mol of copper is completely dissolved in nitric acid, and the average relative molecular weight of the mixed gas of 0.05 mol of nitrogen oxides NO, NO2 and N2O4 can be
A.30 B. 46 C. 50 D. 66
Question 6: Discuss the application method.
Example 6 A certain amount of CO2 gas was slowly introduced into 300 ml KOH solution, and the solution was fully reacted. Evaporating the solution at low temperature and reduced pressure to obtain a white solid. Please answer the following questions:
(1) Because the white solid components obtained by introducing different amounts of carbon dioxide are also different, we try to infer several possibilities and list them.
(2) 2) The CO2 gas is 2.24L (under standard conditions), and the white solid is determined as 1 1.9 g in the white group. What's its mass? Used in many
Coach] (1) Because of the ratio of carbon dioxide and KOH products in the reaction, the amount of KOH solution is related to the concentration of substances, which can be discussed according to the ratio of substances in the reaction. ① When CO2+2 KOH = K 2 CO 3+H2O2+KOH = known N(CO 2)/ potassium hydroxide (KOH)= 1/2, the product affected by K2CO3, N(CO 2)/ N(KOH)= 1, is carbonic acid. 1/2 hours, products with excessive potassium hydroxide are regarded as; 1 N(CO 2)/ KOH, carbon dioxide is excessive, and potassium bicarbonate is affected by solid products. Answer: ① K2CO3+KOH K2CO3+KHCO34KHCO3③ in potassium carbonate.
(2): ① In CO2+2 KOH = K2CO3+H2O2 CO2+KOH = KHCO3.
22.4L (standard state)138g
2.24l (standard state)13.8g 2.24l (standard state)10.0g.
∫13.8g >11.9g1.0g.
∴ The obtained white solid mixture of K2CO3 and KHCO3. BR/>； White solid K2CO3, KHCO3 mole.
① Carbon dioxide+potassium hydroxide = potassium carbonate +H2O② Carbon dioxide+potassium hydroxide = potassium bicarbonate.
X mole, twice mole, x mole, x mole.
Is x mole +Y mole = 2.24L/22.4mol? L- 1 = 0. 100 mol (carbon dioxide)
138g mol- 1x mol 100g？ × y mol of mol- 1 =11.9 (white solid)
Solving this equation, we get
X = 0.0500mol mol (K2CO3)
Y = 0.0500mol mol (potassium bicarbonate)
∴ White solid, K2CO3 mass138g mol-1× 0.0500mol = 6.90g khco3 mass 100g? × 0.0500mol+0 of mol-65438 = 5.00g.
The amount of KOH substance consumed
Twice mole+mole = 2×0.0500 mole 0.0500 mole = 0. 150 mole.
∴ Concentration and dosage of the substance in KOH solution.
0. 150mol/0.300 L = 0.500 0.500mol L- 1。
Summary of law]
Due to the uncertainty of conditions, the results of a class of chemical calculation problems may be two or more, or they may be worth? This practice needs to be solved through discussion to some extent. Common type: 1, discuss the degree of reaction and whether there is excessive reactant; 3. Discuss the composition of reactants or products; Discuss the solution of indefinite equation.
Combined use
6. Pour the water in a large test tube containing the mixed gas of 10 ml NO2 and 10 ml NO at room temperature. When it is slowly transformed into O2 for a period of time, the amount of O2 introduced into the residual gas in vitro will increase in May.
[test work]
1, multiple-choice questions (each question has 12 answers)
1。 In the sulfuric acid solution with the density of 1.45g/cm3, the BaCl2 solution was added drop by drop until it was completely precipitated. It is known that the quality of precipitation is the same as that of original sulfuric acid solution and concentrated sulfuric acid solution.
.29.6% b 42.1%c14.8 mol/l d.6.22 mol/l.
2. In the mixed solution of magnesium chloride, potassium chloride and potassium sulfate, if there are 0.5mol of K+ and Cl- per 1.5mol of Mg2+, the amount of SO42- is 0.5mol.
A. 0. 1 mol b. 0.5 mol
3。 3.20gCu and 30.0mL 10.0mol/L If there is nitric acid in the reaction solution, the product only contains NO and NO2. Molar H+, the amount of NO3- ions contained in the substance solution.
Aa/2mol B.2a according to country c.0. 1a according to country D.(+0. 1) mol >: 4. Under certain conditions, when the reaction of ammonia reaches equilibrium, the volume fraction of ammonia in the mixed gas is 0.25. If the conditions before the reaction are kept unchanged, the gas volume ratio after the reaction is reduced to the original reactant.
a . 1/5 b . 1/4 c . 1/3d . 1/2
5.3 g of magnesium and aluminum alloy only reacted completely with 100 ml of dilute sulfuric acid, and the obtained solution was evaporated to dryness to obtain 17.4 g of anhydrous sulfuric acid.
A. B. 1 mol/L, c. 2 mol/L, d. 2.5 mol/L with the concentration of sulfuric acid solution of the original substance.
6。 Potassium carbonate contains one or two impurities from soda ash, potassium nitrate and barium nitrate. Dissolve 7.8 grams of sample in water to obtain a clear solution, and then add excessive calcium chloride solution to obtain 5.0 grams of impurities in the original sample to confirm the correctness.
A. sodium carbonate and Ba(NO3)2 B. there must be ba (NO3) 2 and sodium carbonate.
C.Ba(NO3)2, and KNO3, and possibly sodium carbonate.
D. When there is no sodium carbonate, but BA(NO3)2 and KNO3 7. 1 are equipped with 2 moles of H 2 O (gas), under certain conditions, the reversible reaction: CO+H 2 O (gas) CO 2+H2, and CO2 0.6 are rubbed to generate reaction equilibrium H 2 O, and under the same conditions, 2 moles of H2O (gas) are rubbed to generate the amount of carbon dioxide.
A. 0.3 mol b. 0.6 mol c. 0.9 mol d. 1.2 mol
8。 The mass of the mixture of KOH and KHCO 3 is 25.6 g. After the mixture is completely calcined in a closed container at 250℃, the residual solid is cooled to 20.7 g in the waste gas. The original mixture has the correct relationship between the amounts of KOH and KHCO3.
A is any ratio greater than b, less than c and equal to d.
9 One of them reacts with 5 mol/L hydrochloric acid 140ML iron oxide sample until it is completely dissolved. The obtained solution can be completely converted into ferrous ions and iron ions under standard conditions, and the oxide with the chemical formula absorbs 0.56L of chlorine 5.
A. Iron oxide b. fe3o 4 C. fe3o 4? Iron pentoxide
10。 The volume ratio of the synthetic tower for synthesizing ammonia with N2 and H2 is 1:3. In the process of NH3 in the synthetic tower, the waste gas mixture accounts for 12%, and the volume fraction of N2 (gas measured at the same temperature and pressure).
A. 12
The calculation topic is 1 1, the pressure of hydrogen and oxygen is 120℃, and the reaction occurs after ignition. When the gas temperature and pressure return to the original measured bmol, the amount of hydrogen and oxygen in the original total gas is calculated.
12, ammonia catalytic oxidation method used in industry can be made from nitric acid, adding dehydrating agent to obtain higher concentration of nitric acid10.7 million tons of synthetic ammonia as raw material, and NH3 becomes HNO3, all assuming.
(1) Write the chemical reaction equation of completely transforming NH3 into HNO3.
(2) In the production process, it is necessary to supplement the absorption and stripping of water. If the dehydrating agent is the same amount of water, what is the mass fraction of nitric acid solution?
(3) If in the production system, 50% nitric acid and 90% nitric acid of m 1 t are both square meters tons (stripping water can be recycled to prepare low-concentration dilute nitric acid), and one * * * has 27 tons of water, then what is the mass ratio of m 1 to m2? BR/>； 13. Existing substances such as sodium bicarbonate and potassium bicarbonate mixed with AG react with 100mL hydrochloric acid. The volume of gas is considered under standard conditions. Fill in the blanks and use the letter formula.
The mass ratio of sodium bicarbonate to KHCO 3 in (1) mixture.
(2) If bicarbonate reacts with hydrochloric acid completely, the substance concentration of hydrochloric acid in (3) is mol L- 1.
If the hydrochloric acid is excessive, the volume of carbon dioxide produced is L.
(4) If the residual hydrochloric acid in bicarbonate is not enough to calculate the volume of CO2, it is necessary to know.
/& gt； (the amount of NaHCO 3 and KHCO 3, not the volume range of carbon dioxide produced when the solid mixture of AG completely reacts with sufficient hydrochloric acid.
14, 1.68L colorless combustible gas is completely burned under standard conditions with enough oxygen. If a sufficient amount of clear limewater is introduced into the product, the quality of white precipitate obtained is 15.0g, and the combustion product absorbs a sufficient amount of 9.3g of alkaline lime. ..
(1) The calculated weight of water in combustion products increases.
(2) If the original gas is a single gas, we can infer br/> by calculating its formula. (3) If the original gas is a mixture of two substances, such as gas, which is the amount of hydrocarbons, please write your own formula. (Only one set is required)
15, a small amount of SO2 in the waste gas is discharged in contact with sulfuric acid to prevent air pollution, and comprehensive utilization is attempted before discharge.
(1) The sulfuric acid plant discharges 0.2% (volume fraction) of ten thousand cubic meters of waste gas, SO 2, which needs to be treated with NaOH solution, lime and O 2. Assuming that elemental sulfur is not lost, the number of kilograms of gypsum (calcium sulfate 2H2O) can be obtained theoretically.
(2) If a certain volume of waste gas is sent to 100mL2mol? The NaOH solution of L- 1 reacted completely, and the obtained solution contained16.7g solute. Test and analyze the composition, and calculate and determine the amount of each component.
(3) In the factory of gypsum waste gas treatment system, the intermediate product of sodium bisulfite is to improve the output by adjusting the ratio of obtaining the optimal flow rate to the amount of waste gas discharged by using NaOH SO2. Now suppose that nNaOH and nnahso 3·NSO 2 represent the amounts of sulfur dioxide, sodium hydroxide and sodium bisulfite, respectively, and are equal to X. Try to write the relationship between the value of X n (sodium bisulfite) in different ranges or the style of sodium bisulfite NSO 2 and nNaOH.
XN (sodium bisulfite)
Reference answer
Test work: 1, BD 2, B 3, D 4, A-5, B,
When 1 1, (1) is completely reacted, the mole of NH 2(AB) and the mole of NO2(AB) are the same;
(2) The remaining time of 2)H2, D, B, C 8 C 7 9 10 NO2, NH2 B molar (AB molar) O2 surplus;
(3), NH 2(AB), NO2(2B) mol.
12，( 1)NH3 +2 O2 = HNO3 + H2O
(2)NH3+2o 2 = = HNO 3+H2O(3)0.5m 1+( 1-0.9)M2 = 27+ 18 & lt； BR/ 17 tons, 63 tons, 18 tons 0.5m 1+0.9m2 = 63.
630,000 tons/year (63 tons 18 tons) ×100% = 77.8% m1/m2 =19/5.
13 (1) 84:100 (2) 5a/46 (3) 22.4a/92 or 5.6a/23(4) hydrochloric acid concentration.
(5)(22.4/ 100) - (22.4/84)
14 (1) 2.7g (2)C 2 H 4(3)C4H6 and H2 or C3H8 and CO, C3H6 and CH 2 O, etc.
15, solution (1) 153.6 kg (2) Author: sodium bisulfite is 0. 1 mol sodium sulfite is 0.05 mol (3)
nNaHSO3
X≤
& ltX & lt 1
nNaHSO3 = 2nSO2 nNaOH
X≥ 1 nNaHSO3 = nNaOH