Problems No. 35 on the Unified State Exam in chemistry
Algorithm for solving such tasks
1. General formula of the homologous series
The most commonly used formulas are summarized in the table:
| Homologous series | General formula |
| Saturated monohydric alcohols | |
| Saturated aldehydes | C n H 2n+1 SON |
| Saturated monocarboxylic acids | C n H 2n+1 COOH |
2. Reaction equation
1) ALL organic substances burn in oxygen to form carbon dioxide, water, nitrogen (if N is present in the compound) and HCl (if chlorine is present):
C n H m O q N x Cl y + O 2 = CO 2 + H 2 O + N 2 + HCl (without coefficients!)
2) Alkenes, alkynes, dienes are prone to addition reactions (reactions with halogens, hydrogen, hydrogen halides, water):
C n H 2n + Cl 2 = C n H 2n Cl 2
C n H 2n + H 2 = C n H 2n+2
C n H 2n + HBr = C n H 2n+1 Br
C n H 2n + H 2 O = C n H 2n+1 OH
Alkynes and dienes, unlike alkenes, add up to 2 moles of hydrogen, chlorine or hydrogen halide per 1 mole of hydrocarbon:
C n H 2n-2 + 2Cl 2 = C n H 2n-2 Cl 4
C n H 2n-2 + 2H 2 = C n H 2n+2
When water is added to alkynes, carbonyl compounds are formed, not alcohols!
3) Alcohols are characterized by reactions of dehydration (intramolecular and intermolecular), oxidation (to carbonyl compounds and, possibly, further to carboxylic acids). Alcohols (including polyhydric ones) react with alkali metals to release hydrogen:
C n H 2n+1 OH = C n H 2n + H 2 O
2C n H 2n+1 OH = C n H 2n+1 OC n H 2n+1 + H 2 O
2C n H 2n+1 OH + 2Na = 2C n H 2n+1 ONa + H 2
4) The chemical properties of aldehydes are very diverse, but here we will only remember redox reactions:
C n H 2n+1 COH + H 2 = C n H 2n+1 CH 2 OH (reduction of carbonyl compounds in the addition of Ni),
C n H 2n+1 COH + [O] = C n H 2n+1 COOH
important point: the oxidation of formaldehyde (HCO) does not stop at the formic acid stage, HCOOH is further oxidized to CO 2 and H 2 O.
5) Carboxylic acids exhibit all the properties of “ordinary” inorganic acids: they interact with bases and basic oxides, react with active metals and salts of weak acids (for example, with carbonates and bicarbonates). The esterification reaction is very important - the formation of esters when interacting with alcohols.
C n H 2n+1 COOH + KOH = C n H 2n+1 COOK + H 2 O
2C n H 2n+1 COOH + CaO = (C n H 2n+1 COO) 2 Ca + H 2 O
2C n H 2n+1 COOH + Mg = (C n H 2n+1 COO) 2 Mg + H 2
C n H 2n+1 COOH + NaHCO 3 = C n H 2n+1 COONa + H 2 O + CO 2
C n H 2n+1 COOH + C 2 H 5 OH = C n H 2n+1 COOC 2 H 5 + H 2 O
3. Finding the amount of a substance by its mass (volume)
formula connecting the mass of a substance (m), its quantity (n) and molar mass (M):
m = n*M or n = m/M.
For example, 710 g of chlorine (Cl 2) corresponds to 710/71 = 10 mol of this substance, since the molar mass of chlorine = 71 g/mol.
For gaseous substances, it is more convenient to work with volumes rather than masses. Let me remind you that the amount of a substance and its volume are related by the following formula: V = V m *n, where V m is the molar volume of the gas (22.4 l/mol under normal conditions).
4. Calculations using reaction equations
This is probably the main type of calculations in chemistry. If you do not feel confident in solving such problems, you need to practice.
The basic idea is this: the quantities of reactants and products formed are related in the same way as the corresponding coefficients in the reaction equation (which is why it is so important to place them correctly!)
Consider, for example, the following reaction: A + 3B = 2C + 5D. The equation shows that 1 mol A and 3 mol B upon interaction form 2 mol C and 5 mol D. The amount of B is three times greater than the amount of substance A, the amount of D is 2.5 times greater than the amount of C, etc. If in If the reaction is not 1 mol A, but, say, 10, then the amounts of all other participants in the reaction will increase exactly 10 times: 30 mol B, 20 mol C, 50 mol D. If we know that 15 mol D were formed (three times times more than indicated in the equation), then the amounts of all other compounds will be 3 times greater.
5. Calculation of the molar mass of the test substance
The mass X is usually given in the problem statement; we found the quantity X in paragraph 4. It remains to use the formula M = m/n again.
6. Determination of the molecular formula of X.
The final stage. Knowing the molar mass of X and the general formula of the corresponding homologous series, you can find the molecular formula of the unknown substance.
Let, for example, the relative molecular weight of the limiting monohydric alcohol be 46. The general formula of the homologous series: C n H 2n+1 OH. Relative molecular weight consists of the mass of n carbon atoms, 2n+2 hydrogen atoms and one oxygen atom. We get the equation: 12n + 2n + 2 + 16 = 46. Solving the equation, we find that n = 2. The molecular formula of alcohol is: C 2 H 5 OH.
Don't forget to write down your answer!
Example 1 . 10.5 g of some alkene can add 40 g of bromine. Identify the unknown alkene.
Solution. Let a molecule of an unknown alkene contain n carbon atoms. General formula of the homologous series C n H 2n. Alkenes react with bromine according to the equation:
CnH2n + Br2 = CnH2nBr2.
Let's calculate the amount of bromine that entered the reaction: M(Br 2) = 160 g/mol. n(Br 2) = m/M = 40/160 = 0.25 mol.
The equation shows that 1 mol of alkene adds 1 mol of bromine, therefore, n(C n H 2n) = n(Br 2) = 0.25 mol.
Knowing the mass of the reacted alkene and its quantity, we will find its molar mass: M(C n H 2n) = m(mass)/n(amount) = 10.5/0.25 = 42 (g/mol).
Now it is quite easy to identify an alkene: the relative molecular weight (42) is the sum of the mass of n carbon atoms and 2n hydrogen atoms. We get the simplest algebraic equation:
The solution to this equation is n = 3. The alkene formula is: C 3 H 6 .
Answer: C 3 H 6 .
Example 2 . The complete hydrogenation of 5.4 g of some alkyne requires 4.48 liters of hydrogen (n.s.). Determine the molecular formula of this alkyne.
Solution. We will act in accordance with the general plan. Let a molecule of an unknown alkyne contain n carbon atoms. General formula of the homologous series C n H 2n-2. The hydrogenation of alkynes proceeds according to the equation:
C n H 2n-2 + 2H 2 = C n H 2n+2.
The amount of hydrogen that reacted can be found using the formula n = V/Vm. In this case, n = 4.48/22.4 = 0.2 mol.
The equation shows that 1 mol of alkyne adds 2 mol of hydrogen (remember that the problem statement refers to complete hydrogenation), therefore, n(C n H 2n-2) = 0.1 mol.
Based on the mass and amount of the alkyne, we find its molar mass: M(C n H 2n-2) = m(mass)/n(amount) = 5.4/0.1 = 54 (g/mol).
The relative molecular weight of an alkyne is the sum of n atomic masses of carbon and 2n-2 atomic masses of hydrogen. We get the equation:
12n + 2n - 2 = 54.
We solve the linear equation, we get: n = 4. Alkyne formula: C 4 H 6.
Answer: C 4 H 6 .
Example 3 . When 112 liters (n.a.) of an unknown cycloalkane are burned in excess oxygen, 336 liters of CO 2 are formed. Establish the structural formula of the cycloalkane.
Solution. The general formula of the homologous series of cycloalkanes: C n H 2n. With complete combustion of cycloalkanes, as with the combustion of any hydrocarbons, carbon dioxide and water are formed:
C n H 2n + 1.5n O 2 = n CO 2 + n H 2 O.
Please note: the coefficients in the reaction equation in this case depend on n!
During the reaction, 336/22.4 = 15 moles of carbon dioxide were formed. 112/22.4 = 5 moles of hydrocarbon entered the reaction.
Further reasoning is obvious: if 15 moles of CO 2 are formed per 5 moles of cycloalkane, then 15 molecules of carbon dioxide are formed per 5 molecules of hydrocarbon, i.e., one cycloalkane molecule produces 3 CO 2 molecules. Since each molecule of carbon monoxide (IV) contains one carbon atom, we can conclude: one cycloalkane molecule contains 3 carbon atoms.
Conclusion: n = 3, cycloalkane formula - C 3 H 6.
The formula C 3 H 6 corresponds to only one isomer - cyclopropane.
Answer: cyclopropane.
Example 4 . 116 g of some saturated aldehyde were heated for a long time with an ammonia solution of silver oxide. The reaction produced 432 g of metallic silver. Determine the molecular formula of the aldehyde.
Solution. The general formula of the homologous series of saturated aldehydes is: C n H 2n+1 COH. Aldehydes are easily oxidized to carboxylic acids, in particular, under the action of an ammonia solution of silver oxide:
C n H 2n+1 COH + Ag 2 O = C n H 2n+1 COOH + 2Ag.
Note. In reality, the reaction is described by a more complex equation. When Ag 2 O is added to an aqueous ammonia solution, a complex compound OH is formed - diammine silver hydroxide. It is this compound that acts as an oxidizing agent. During the reaction, an ammonium salt of a carboxylic acid is formed:
C n H 2n+1 COH + 2OH = C n H 2n+1 COONH 4 + 2Ag + 3NH 3 + H 2 O.
Another important point! The oxidation of formaldehyde (HCOH) is not described by the given equation. When HCOH reacts with an ammonia solution of silver oxide, 4 moles of Ag per 1 mole of aldehyde are released:
НCOH + 2Ag2O = CO2 + H2O + 4Ag.
Be careful when solving problems involving the oxidation of carbonyl compounds!
Let's return to our example. Based on the mass of released silver, you can find the amount of this metal: n(Ag) = m/M = 432/108 = 4 (mol). In accordance with the equation, 2 moles of silver are formed per 1 mole of aldehyde, therefore, n(aldehyde) = 0.5n(Ag) = 0.5*4 = 2 moles.
Molar mass of aldehyde = 116/2 = 58 g/mol. Try to do the next steps yourself: you need to create an equation, solve it and draw conclusions.
Answer: C 2 H 5 COH.
Example 5 . When 3.1 g of a certain primary amine reacts with a sufficient amount of HBr, 11.2 g of salt is formed. Determine the formula of the amine.
Solution. Primary amines (C n H 2n + 1 NH 2) when reacting with acids form alkylammonium salts:
С n H 2n+1 NH 2 + HBr = [С n H 2n+1 NH 3 ] + Br - .
Unfortunately, based on the mass of the amine and the salt formed, we will not be able to find their quantities (since the molar masses are unknown). Let's take a different path. Let us remember the law of conservation of mass: m(amine) + m(HBr) = m(salt), therefore, m(HBr) = m(salt) - m(amine) = 11.2 - 3.1 = 8.1.
Pay attention to this technique, which is very often used when solving C 5. Even if the mass of the reagent is not given explicitly in the problem statement, you can try to find it from the masses of other compounds.
So, we are back on track with the standard algorithm. Based on the mass of hydrogen bromide, we find the amount, n(HBr) = n(amine), M(amine) = 31 g/mol.
Answer: CH 3 NH 2 .
Example 6 . A certain amount of alkene X, when reacting with an excess of chlorine, forms 11.3 g of dichloride, and when reacting with an excess of bromine, 20.2 g of dibromide. Determine the molecular formula of X.
Solution. Alkenes add chlorine and bromine to form dihalogen derivatives:
C n H 2n + Cl 2 = C n H 2n Cl 2,
C n H 2n + Br 2 = C n H 2n Br 2.
In this problem it is pointless to try to find the amount of dichloride or dibromide (their molar masses are unknown) or the amount of chlorine or bromine (their masses are unknown).
We use one non-standard technique. The molar mass of C n H 2n Cl 2 is 12n + 2n + 71 = 14n + 71. M(C n H 2n Br 2) = 14n + 160.
The masses of dihalides are also known. You can find the amounts of substances obtained: n(C n H 2n Cl 2) = m/M = 11.3/(14n + 71). n(C n H 2n Br 2) = 20.2/(14n + 160).
By convention, the amount of dichloride is equal to the amount of dibromide. This fact allows us to create the equation: 11.3/(14n + 71) = 20.2/(14n + 160).
This equation has a unique solution: n = 3.
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Option No. 1380120
Tasks 34 (C5). Sergey Shirokopoyas: Chemistry - preparation for the Unified State Exam 2016
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Some or-ga-no-substance A contains by mass 11.97% nitrogen, 51.28% carbon-le-ro-da, 27.35 % sour, and water. A is formed by the interaction of substance B with pro-pa-no-lom-2 in a mo-lar co-from-no-she- Research 1: 1. It is known that substance B has a natural origin.
1) About the calculations that are not necessary to find the formula of substance A;
2) Establish its mo-le-ku-lyar-nu-lu-lu;
3) Create a structural form of substance A, which creates a series of connections between atoms in the mo- le-ku-le;
4) Write down the equation for the reaction of substance A from substance B and pro-pa-no-la-2.
Upon combustion of 40.95 g of organic matter, 39.2 liters of carbon dioxide (n.o.), 3.92 liters of nitrogen (n.o.) and 34.65 g of water were obtained. When heated with hydrochloric acid, this substance undergoes hydrolysis, the products of which are compounds of the composition and secondary alcohol.
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The primary amine salt reacted with silver nitrate, resulting in a precipitate and the formation of organic substance A, containing by weight 29.79% nitrogen, 51.06% oxygen and 12.77% carbon.
Based on the data of the problem conditions:
2) establish its molecular formula;
3) create a structural formula of this substance A, which reflects the order of bonds of atoms in the molecule;
4) write the equation for the reaction of obtaining substance A from the salt of the primary amine and.
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When burning a dipeptide of natural origin weighing 2.64 g, 1.792 liters of carbon dioxide (n.s.), 1.44 g of water and 448 ml of nitrogen (n.s.) were obtained. When this substance was hydrolyzed in the presence of hydrochloric acid, only one salt was formed.
Based on the data of the problem conditions:
2) establish its molecular formula;
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Some organic substance A contains by weight 13.58% nitrogen, 46.59% carbon and 31.03% oxygen and is formed by the interaction of substance B with ethanol in a 1:1 molar ratio. It is known that substance B is of natural origin.
Based on the data of the problem conditions:
1) make the calculations necessary to find the formula of substance A;
2) establish its molecular formula;
3) create a structural formula of substance A, which reflects the order of bonds of atoms in the molecule;
4) write the equation for the reaction of obtaining substance A from substance B and ethanol.
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Some organic substance A contains by mass 10.68% nitrogen, 54.94% carbon and 24.39 % acidity and is formed during the interaction of substance B with prop-no-lom-1 in mo-lar from-no-she-nii 1: 1. It is known that substance B is a natural ami-no-acid.
Based on the given conditions:
1) about the calculations that are not needed to find the formula of substance A;
2) establish its molecular form;
3) create a structural form of substance A, which creates a series of connections between atoms in the mo- le-ku-le;
4) write the equation for the reaction of obtaining substance A from substance B and n-pro-pa-no-la.
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A certain substance, which is a salt of organic origin, contains by weight 12.79% nitrogen, 43.84% carbon and 32.42% chlorine and is formed by the reaction of a primary amine with chloroethane.
Based on the data of the problem conditions:
1) make the calculations necessary to find the formula of the original organic substance;
2) establish its molecular formula;
3) create a structural formula of this substance, which reflects the order of bonds of atoms in the molecule;
4) write the reaction equation for producing this substance from primary amine and chloroethane.
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When burning a dipeptide of natural origin weighing 3.2 g, 2.688 liters of carbon dioxide (n.s.), 448 ml of nitrogen (n.s.) and 2.16 g of water were obtained. When this substance was hydrolyzed in the presence of potassium hydroxide, only one salt was formed.
Based on the data of the problem conditions:
1) make the calculations necessary to find the formula of the dipeptide;
2) establish its molecular formula;
3) create a structural formula of the dipeptide, which reflects the order of bonds of atoms in the molecule;
4) write the reaction equation for the hydrolysis of this dipeptide in the presence of potassium hydroxide.
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When burning a dipeptide of natural origin weighing 6.4 g, 5.376 liters of carbon dioxide (n.s.), 896 ml of nitrogen (n.s.) and 4.32 g of water were obtained. When this substance was hydrolyzed in the presence of hydrochloric acid, only one salt was formed.
Based on the data of the problem conditions:
1) make the calculations necessary to find the formula of the dipeptide;
2) establish its molecular formula;
3) create a structural formula of the dipeptide, which reflects the order of bonds of atoms in the molecule;
4) write the reaction equation for the hydrolysis of this dipeptide in the presence of hydrochloric acid.
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The combustion of some organic substance weighing 4.12 g produced 3.584 liters of carbon dioxide (n.s.), 448 ml of nitrogen (n.s.) and 3.24 g of water. When heated with hydrochloric acid, this substance undergoes hydrolysis, the products of which are compounds of the composition and alcohol.
Based on the data of the problem conditions:
1) make the calculations necessary to find the formula of the original organic substance;
2) establish its molecular formula;
3) create a structural formula of this substance, which reflects the order of bonds of atoms in the molecule;
4) write the equation for the hydrolysis reaction of this substance in the presence of hydrochloric acid.
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When a certain organic substance weighing 4.68 g was burned, 4.48 liters of carbon dioxide (n.s.), 448 ml of nitrogen (n.s.) and 3.96 g of water were obtained. When heated with a solution of sodium hydroxide, this substance undergoes hydrolysis, the products of which are a salt of a natural amino acid and a secondary alcohol.
Based on the data of the problem conditions:
1) make the calculations necessary to find the formula of the original organic substance;
2) establish its molecular formula;
3) create a structural formula of this substance, which reflects the order of bonds of atoms in the molecule;
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When a certain organic substance weighing 17.55 g was burned, 16.8 liters of carbon dioxide (n.s.), 1.68 liters of nitrogen (n.s.) and 14.85 g of water were obtained. When heated with a solution of sodium hydroxide, this substance undergoes hydrolysis, the products of which are a salt of a natural amino acid and a secondary alcohol.
Based on the data of the problem conditions:
1) make the calculations necessary to find the formula of the original organic substance;
2) establish its molecular formula;
3) create a structural formula of this substance, which reflects the order of bonds of atoms in the molecule;
4) write the equation for the hydrolysis reaction of this substance in the presence of sodium hydroxide.
Solutions to long-answer tasks are not automatically checked.
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When a certain organic substance weighing 35.1 g was burned, 33.6 liters of carbon dioxide (n.s.), 3.36 liters of nitrogen (n.s.) and 29.7 g of water were obtained. When heated with a solution of potassium hydroxide, this substance undergoes hydrolysis, the products of which are a salt of a natural amino acid and a secondary alcohol.
Solution 34 of the Unified State Exam 2018 task in chemistry from the demo version. Checked content elements: Calculations of the mass (volume, amount of substance) of reaction products, if one of the substances is given in excess (has impurities). Calculations using the concept of “mass fraction of a substance in solution”. Calculations of the mass or volume fraction of the yield of the reaction product from the theoretically possible. Calculation of the mass fraction (mass) of a chemical compound in a mixture.
When a sample of calcium carbonate was heated, some of the substance decomposed. At the same time, 4.48 liters (n.s.) of carbon dioxide were released. The mass of the solid residue was 41.2 g. This residue was added to 465.5 g of hydrochloric acid solution, taken in excess. Determine the mass fraction of salt in the resulting solution.
In your answer, write down the reaction equations that are indicated in the problem statement and provide all the necessary calculations (indicate the units of measurement of the required physical quantities).
Solution 34 of the Unified State Exam 2018 task in chemistry
The reaction equations are written:
CaCO 3 = CaO + CO 2
CaCO 3 + 2HCl = CaCl 2 + CO 2 + H 2 O
CaO + 2HCl = CaCl 2 + H 2 O
The amount of substance compounds in the solid residue was calculated:
n(CO 2) = V / V m = 4.48 / 22.4 = 0.2 mol
n(CaO) = n(CO 2) = 0.2 mol
m(CaO) = n M = 0.2 56 = 11.2 g
m(CaCO 3 residue) = 41.2 – 11.2 = 30 g
n(CaCO 3 residue) = m / M = 30 / 100 = 0.3 mol
The mass of salt in the resulting solution was calculated:
n(CaCl 2) = n(CaO) + n(CaCO 3) = 0.5 mol
m(CaCl 2) = n M = 0.5 111 = 55.5 g
n(CO 2) = n(CaCO 3 residue) = 0.3 mol
m(CO 2) = n M = 0.3 44 = 13.2 g
The mass fraction of calcium chloride in the solution is calculated:
m(solution) = 41.2 + 465.5 – 13.2 = 493.5 g
ω(CaCl 2) = m(CaCl 2) / m(solution) = 55.5 / 493.5 = 0.112, or 11.2%
Task 34
When a sample of calcium carbonate was heated, some of the substance decomposed. At the same time, 4.48 liters (n.s.) of carbon dioxide were released. The mass of the solid residue was 41.2 g. This residue was added to 465.5 g of a solution of hydrochloric acid taken in excess. Determine the mass fraction of salt in the resulting solution.
In your answer, write down the reaction equations that are indicated in the problem statement and provide all the necessary calculations (indicate the units of measurement of the required quantities).
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Answer: Let us write down a brief condition for this problem.
After all the preparations have been made, we proceed to the decision.
1) Determine the amount of CO 2 contained in 4.48 liters. his.
n(CO 2) = V/Vm = 4.48 l / 22.4 l/mol = 0.2 mol
2) Determine the amount of calcium oxide formed.
According to the reaction equation, 1 mol CO 2 and 1 mol CaO are formed
Hence: n(CO2) = n(CaO) and equals 0.2 mol
3) Determine the mass of 0.2 mol CaO
m(CaO) = n(CaO) M(CaO) = 0.2 mol 56 g/mol = 11.2 g
Thus, a solid residue weighing 41.2 g consists of 11.2 g of CaO and (41.2 g - 11.2 g) 30 g of CaCO 3
4) Determine the amount of CaCO 3 contained in 30 g
n(CaCO3) = m(CaCO 3) / M(CaCO 3) = 30 g / 100 g/mol = 0.3 mol
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CaO + HCl = CaCl 2 + H 2 O
CaCO 3 + HCl = CaCl 2 + H 2 O + CO 2
5) Determine the amount of calcium chloride formed as a result of these reactions.
The reaction involved 0.3 mol of CaCO 3 and 0.2 mol of CaO for a total of 0.5 mol.
Accordingly, 0.5 mol CaCl 2 is formed
6) Calculate the mass of 0.5 mol calcium chloride
M(CaCl2) = n(CaCl2) M(CaCl 2) = 0.5 mol · 111 g/mol = 55.5 g.
7) Determine the mass of carbon dioxide. The decomposition reaction involved 0.3 mol of calcium carbonate, therefore:
n(CaCO3) = n(CO 2) = 0.3 mol,
m(CO2) = n(CO2) M(CO 2) = 0.3 mol · 44 g/mol = 13.2 g.
8) Find the mass of the solution. It consists of the mass of hydrochloric acid + the mass of the solid residue (CaCO 3 + CaO) minutes, the mass of the released CO 2. Let's write this as a formula:
m(r-ra) = m(CaCO 3 + CaO) + m(HCl) – m(CO 2) = 465.5 g + 41.2 g – 13.2 g = 493.5 g.
The new reference book contains all the theoretical material on the chemistry course required to pass the Unified State Exam. It includes all elements of content, verified by test materials, and helps to generalize and systematize knowledge and skills for a secondary (high) school course. Theoretical material is presented in a concise, accessible form. Each section is accompanied by examples of training tasks that allow you to test your knowledge and degree of preparedness for the certification exam. Practical tasks correspond to the Unified State Exam format. At the end of the manual, answers to tasks are provided that will help you objectively assess the level of your knowledge and the degree of preparedness for the certification exam. The manual is addressed to high school students, applicants and teachers.
9) And finally, we will answer the question of the task. Let's find the mass fraction in % of salt in the solution using the following magic triangle:
ω%(CaCI 2) = m(CaCI 2) / m(solution) = 55.5 g / 493.5 g = 0.112 or 11.2%
Answer: ω% (CaCI 2) = 11.2%