Exam chemistry online tests with scores. Preparing for the Unified State Exam in Chemistry

27.09.2019

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Greetings to schoolchildren who have entered 11th grade! The last school year is the most memorable and important in a student’s life. After all, you need to finally make a choice future specialty and subjects for exams. This time I have chosen for you useful material about how to prepare for the Unified State Exam in chemistry.

Theory for preparing for the Unified State Exam in chemistry

Exam preparation always begins with studying the theoretical part. Therefore, if your knowledge of chemistry is at an average level, improve the theory, but reinforce it with practical exercises.

In 2018, the Unified State Exam in Chemistry consisted of 35 tasks: the first 29 questions require choosing an answer from those proposed, or writing a digital answer after calculation, the remaining 6 tasks require providing a complete, detailed answer. For the first 29 answers you can score a maximum of 40 points, and for the second part of the Unified State Exam - 20 points. Perhaps in 2019 the structure of the Unified State Exam in chemistry will remain unchanged.

The main theoretical questions in the Unified State Exam in chemistry cover the following topics:

The structure of the atom in the modern understanding.
Mendeleev table.
Inorganic chemistry (chemical properties of metals and non-metals).
Organic chemistry (fats, proteins and carbohydrates).
Experimental chemistry in theory (working and safety rules in the laboratory, methods for obtaining a certain substance).
Ideas about methods for obtaining the required substances and elements in industrial version(metallurgy and methods for producing metals in production, chemical industry).
Calculations using formulas and chemical equations.

Plan preparation for the Unified State Exam in chemistry

1). Create an annual plan with hourly calculations and a choice of preparation days. For example, study chemistry for 2 hours a day on Mondays, Wednesdays and Saturdays.

2). It is best to involve in preparation loved one(parents or sister/brother). If this is not possible, then team up with another student planning to take the Unified State Exam in chemistry. This way you will feel each other’s support and at the same time push you on if one of you is lagging behind. This is a unique way of motivation, and classes will be more interesting.

3). Calculate the time to complete each test task. This way you will know in advance how much time to spend on a question, and if you get stuck on something, you can move on to another task and return to the unfinished task later.

4). As you approach the exam, try to maximize your nutrition and sleep. The examinee should feel rested.

Advice! During the exam itself, you need to decide on the difficulty of the tasks. The items that are easiest for you to understand are best left for the last 30 minutes of the exam. The problems in the second part will give you a high score, so it is recommended to start with them, but it is advisable to adhere to the planned time for completing each task. Answers to simple questions can be given at the end of the exam.

Books for preparing for the Unified State Exam in chemistry

On one's own prepare for the chemistry exam can be done by studying textbooks and methodological manuals. This method is the most difficult, since the student will need maximum concentration, the ability to independently understand the material, perseverance and self-discipline.

Among the popular textbooks for preparing for the Unified State Exam in chemistry are:

“Unified State Exam. Chemistry. Big reference book"(authors - Doronkin, Sazhneva, Berezhnaya). The book describes in detail the main sections of organic and inorganic chemistry, as well as general chemistry. The manual contains tasks for the practical part. The book contains 560 pages. Approximate cost is about 300 rubles.
« Chemistry tutor"(author - Egorov). The book was created for in-depth study of chemistry in preparation for the Unified State Exam. "Tutor" consists of theoretical issues and answers to them (thematic testing), as well as from practical tasks according to difficulty levels with detailed explanation solution algorithm. The book contains 762 pages. Approximate cost is about 600 rubles.

Courses in chemistry: preparation for the Unified State Exam

The most popular and in a simple way preparation for the Unified State Exam in chemistry Attendance at group courses or individual tutoring is accepted. Self-discipline and independent analysis of materials are not required here. A chemistry teacher will set up a time for a visit and help you understand simple and complex tasks ah within the framework of the approved program.

The material provided in chemistry courses is usually based on questions and topics from last year's USE exams. The teacher takes into account the most common mistakes students and gives full analysis such tasks.

Chemistry website for preparing for the Unified State Exam

Distance learning is popular now, so you can take advantage of the opportunity to prepare for the Unified State Exam in chemistry using online lessons. Some of them are free, some are completely paid, and there are online lessons with partial payment, i.e. you can watch the first lesson for free, and then decide to continue training on a paid basis.

State final certification 2019 in chemistry for 9th grade graduates educational institutions is carried out to assess the level of general education training of graduates in this discipline. The tasks test knowledge of the following sections of chemistry:

The structure of the atom.
Periodic law and Periodic table chemical elements D.I. Mendeleev.
The structure of molecules. Chemical bond: covalent (polar and non-polar), ionic, metallic.
Valency of chemical elements. The degree of oxidation of chemical elements.
Simple and complex substances.
Chemical reaction. Conditions and signs of chemical reactions. Chemical equations.
Electrolytes and non-electrolytes. Cations and anions. Electrolytic dissociation acids, alkalis and salts (average).
Ion exchange reactions and conditions for their implementation.
Chemical properties simple substances: metals and non-metals.
Chemical properties of oxides: basic, amphoteric, acidic.
Chemical properties of bases. Chemical properties of acids.
Chemical properties of salts (average).
Pure substances and mixtures. Rules safe work in the school laboratory. Chemical pollution environment and its consequences.
The degree of oxidation of chemical elements. Oxidizing agent and reducing agent. Redox reactions.
Calculation of mass fraction chemical element in matter.
Periodic law D.I. Mendeleev.
Initial information about organic substances. Biologically important substances: proteins fats carbohydrates.
Determination of the nature of the solution environment of acids and alkalis using indicators. Qualitative reactions to ions in solution (chloride, sulfate, carbonation, ammonium ion). Qualitative reactions to gaseous substances (oxygen, hydrogen, carbon dioxide, ammonia).
Chemical properties of simple substances. Chemical properties of complex substances.

Date of passing the OGE in chemistry 2019:
June 4 (Tuesday).

There are no changes in the structure and content of the 2019 examination paper compared to 2018.

In this section you will find online tests that will help you prepare for taking the OGE (GIA) in chemistry. We wish you success!

Standard OGE test(GIA-9) format of 2019 in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. In this regard, in this test Only the first part is presented (i.e. the first 19 tasks). According to the current exam structure, among these tasks, answer options are offered only in 15. However, for the convenience of passing tests, the site administration decided to offer answer options in all tasks. But for tasks in which the compilers of real test and measurement materials (CMMs) do not provide answer options, the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard OGE test (GIA-9) of the 2019 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. In this regard, only the first part (i.e., the first 19 tasks) is presented in this test. According to the current exam structure, among these tasks, answer options are offered only in 15. However, for the convenience of passing tests, the site administration decided to offer answer options in all tasks. But for tasks in which the compilers of real test and measurement materials (CMMs) do not provide answer options, the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard OGE test (GIA-9) of the 2018 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. In this regard, only the first part (i.e., the first 19 tasks) is presented in this test. According to the current exam structure, among these tasks, answer options are offered only in 15. However, for the convenience of passing tests, the site administration decided to offer answer options in all tasks. But for tasks in which the compilers of real test and measurement materials (CMMs) do not provide answer options, the number of answer options has been significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard OGE test (GIA-9) of the 2017 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. In this regard, only the first part (i.e., the first 19 tasks) is presented in this test. According to the current exam structure, among these tasks, answer options are offered only in 15. However, for the convenience of passing tests, the site administration decided to offer answer options in all tasks. But for tasks in which the compilers of real test and measurement materials (CMMs) do not provide answer options, the number of answer options was significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard OGE test (GIA-9) of the 2016 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. In this regard, only the first part (i.e., the first 19 tasks) is presented in this test. According to the current exam structure, among these tasks, answer options are offered only in 15. However, for the convenience of passing tests, the site administration decided to offer answer options in all tasks. But for tasks in which the compilers of real test and measurement materials (CMMs) do not provide answer options, the number of answer options was significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

The standard OGE test (GIA-9) of the 2015 format in chemistry consists of two parts. The first part contains 19 tasks with a short answer, the second part contains 3 tasks with a detailed answer. In this regard, only the first part (i.e., the first 19 tasks) is presented in this test. According to the current exam structure, among these tasks, answer options are offered only in 15. However, for the convenience of passing tests, the site administration decided to offer answer options in all tasks. But for tasks in which the compilers of real test and measurement materials (CMMs) do not provide answer options, the number of answer options was significantly increased in order to bring our test as close as possible to what you will have to face at the end of the school year.

When completing tasks A1-A19, select only one correct option .
When completing tasks B1-B3, select two correct options.

When completing tasks A1-A15, select only one correct option.

When completing tasks A1-A15, choose only one correct option.

Methods for solving problems in chemistry

When solving problems, you must be guided by a few simple rules:

Read the task conditions carefully;
Write down what is given;
Convert, if necessary, units of physical quantities into SI units (some non-system units are allowed, for example liters);
Write down, if necessary, the reaction equation and arrange the coefficients;
Solve a problem using the concept of the amount of a substance, and not the method of drawing up proportions;
Write down the answer.

In order to successful preparation in chemistry, you should carefully consider the solutions to the problems given in the text, and also solve a sufficient number of them yourself. It is in the process of solving problems that the basic theoretical principles of the chemistry course will be reinforced. It is necessary to solve problems throughout the entire time of studying chemistry and preparing for the exam.

You can use the problems on this page, or you can download a good collection of problems and exercises with the solution of standard and complicated problems (M. I. Lebedeva, I. A. Ankudimova): download.

Mole, molar mass

Molar mass is the ratio of the mass of a substance to the amount of substance, i.e.

M(x) = m(x)/ν(x), (1)

where M(x) is the molar mass of substance X, m(x) is the mass of substance X, ν(x) is the amount of substance X. The SI unit of molar mass is kg/mol, but the unit g/mol is usually used. Unit of mass – g, kg. The SI unit for quantity of a substance is the mole.

Any chemistry problem solved through the amount of substance. You need to remember the basic formula:

ν(x) = m(x)/ M(x) = V(x)/V m = N/N A , (2)

where V(x) is the volume of the substance X(l), V m is the molar volume of the gas (l/mol), N is the number of particles, N A is Avogadro’s constant.

1. Determine mass sodium iodide NaI amount of substance 0.6 mol.

Given: ν(NaI)= 0.6 mol.

Find: m(NaI) =?

Solution. The molar mass of sodium iodide is:

M(NaI) = M(Na) + M(I) = 23 + 127 = 150 g/mol

Determine the mass of NaI:

m(NaI) = ν(NaI) M(NaI) = 0.6 150 = 90 g.

2. Determine the amount of substance atomic boron contained in sodium tetraborate Na 2 B 4 O 7 weighing 40.4 g.

Given: m(Na 2 B 4 O 7) = 40.4 g.

Find: ν(B)=?

Solution. The molar mass of sodium tetraborate is 202 g/mol. Determine the amount of substance Na 2 B 4 O 7:

ν(Na 2 B 4 O 7) = m(Na 2 B 4 O 7)/ M(Na 2 B 4 O 7) = 40.4/202 = 0.2 mol.

Recall that 1 mole of sodium tetraborate molecule contains 2 moles of sodium atoms, 4 moles of boron atoms and 7 moles of oxygen atoms (see sodium tetraborate formula). Then the amount of atomic boron substance is equal to: ν(B) = 4 ν (Na 2 B 4 O 7) = 4 0.2 = 0.8 mol.

Calculations according to chemical formulas. Mass fraction.

Mass fraction of a substance is the ratio of the mass of a given substance in a system to the mass of the entire system, i.e. ω(X) =m(X)/m, where ω(X) is the mass fraction of substance X, m(X) is the mass of substance X, m is the mass of the entire system. Mass fraction is a dimensionless quantity. It is expressed as a fraction of a unit or as a percentage. For example, the mass fraction of atomic oxygen is 0.42, or 42%, i.e. ω(O)=0.42. The mass fraction of atomic chlorine in sodium chloride is 0.607, or 60.7%, i.e. ω(Cl)=0.607.

3. Determine the mass fraction water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.

Solution: The molar mass of BaCl 2 2H 2 O is:

M(BaCl 2 2H 2 O) = 137+ 2 35.5 + 2 18 = 244 g/mol

From the formula BaCl 2 2H 2 O it follows that 1 mol of barium chloride dihydrate contains 2 mol of H 2 O. From this we can determine the mass of water contained in BaCl 2 2H 2 O:

m(H 2 O) = 2 18 = 36 g.

Find the mass fraction of water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.

ω(H 2 O) = m(H 2 O)/ m(BaCl 2 2H 2 O) = 36/244 = 0.1475 = 14.75%.

4. From sample rock weighing 25 g, containing the mineral argentite Ag 2 S, silver weighing 5.4 g was isolated. Determine the mass fraction argentite in the sample.

Given: m(Ag)=5.4 g; m = 25 g.

Find: ω(Ag 2 S) =?

Solution: we determine the amount of silver substance found in argentite: ν(Ag) =m(Ag)/M(Ag) = 5.4/108 = 0.05 mol.

From the formula Ag 2 S it follows that the amount of argentite substance is half as much as the amount of silver substance. Determine the amount of argentite substance:

ν(Ag 2 S)= 0.5 ν(Ag) = 0.5 0.05 = 0.025 mol

We calculate the mass of argentite:

m(Ag 2 S) = ν(Ag 2 S) M(Ag 2 S) = 0.025 248 = 6.2 g.

Now we determine the mass fraction of argentite in a rock sample weighing 25 g.

ω(Ag 2 S) = m(Ag 2 S)/ m = 6.2/25 = 0.248 = 24.8%.

Deriving compound formulas

5. Determine the simplest formula of the compound potassium with manganese and oxygen, if the mass fractions of elements in this substance are 24.7, 34.8 and 40.5%, respectively.

Given: ω(K) =24.7%; ω(Mn) =34.8%; ω(O) =40.5%.

Find: formula of the compound.

Solution: for calculations we select the mass of the compound equal to 100 g, i.e. m=100 g. The masses of potassium, manganese and oxygen will be:

m (K) = m ω(K); m (K) = 100 0.247 = 24.7 g;

m (Mn) = m ω(Mn); m (Mn) =100 0.348=34.8 g;

m (O) = m ω(O); m(O) = 100 0.405 = 40.5 g.

We determine the amounts of atomic substances potassium, manganese and oxygen:

ν(K)= m(K)/ M(K) = 24.7/39= 0.63 mol

ν(Mn)= m(Mn)/ М(Mn) = 34.8/ 55 = 0.63 mol

ν(O)= m(O)/ M(O) = 40.5/16 = 2.5 mol

We find the ratio of the quantities of substances:

ν(K) : ν(Mn) : ν(O) = 0.63: 0.63: 2.5.

Dividing the right side of the equality by a smaller number (0.63) we get:

ν(K) : ν(Mn) : ν(O) = 1: 1: 4.

Hence, simplest formula KMnO 4 compounds.

6. The combustion of 1.3 g of a substance produced 4.4 g of carbon monoxide (IV) and 0.9 g of water. Find the molecular formula substance if its hydrogen density is 39.

Given: m(in-va) =1.3 g; m(CO 2)=4.4 g; m(H 2 O) = 0.9 g; D H2 =39.

Find: formula of a substance.

Solution: Let's assume that the substance we are looking for contains carbon, hydrogen and oxygen, because during its combustion, CO 2 and H 2 O were formed. Then it is necessary to find the amounts of CO 2 and H 2 O substances in order to determine the amounts of atomic carbon, hydrogen and oxygen substances.

ν(CO 2) = m(CO 2)/ M(CO 2) = 4.4/44 = 0.1 mol;

ν(H 2 O) = m(H 2 O)/ M(H 2 O) = 0.9/18 = 0.05 mol.

We determine the amounts of atomic carbon and hydrogen substances:

ν(C)= ν(CO 2); ν(C)=0.1 mol;

ν(H)= 2 ν(H 2 O); ν(H) = 2 0.05 = 0.1 mol.

Therefore, the masses of carbon and hydrogen will be equal:

m(C) = ν(C) M(C) = 0.1 12 = 1.2 g;

m(N) = ν(N) M(N) = 0.1 1 =0.1 g.

We define high-quality composition substances:

m(in-va) = m(C) + m(H) = 1.2 + 0.1 = 1.3 g.

Consequently, the substance consists only of carbon and hydrogen (see the problem statement). Let us now determine its molecular weight based on the given condition tasks hydrogen density of a substance.

M(v-va) = 2 D H2 = 2 39 = 78 g/mol.

ν(С) : ν(Н) = 0.1: 0.1

Dividing the right side of the equality by the number 0.1, we get:

ν(С) : ν(Н) = 1: 1

Let us take the number of carbon (or hydrogen) atoms as “x”, then multiply “x” by the atomic masses of carbon and hydrogen and equate this sum molecular weight substances, solve the equation:

12x + x = 78. Hence x = 6. Therefore, the formula of the substance is C 6 H 6 - benzene.

Molar volume of gases. Laws of ideal gases. Volume fraction.

Molar volume of gas equal to the ratio volume of gas to the amount of substance of this gas, i.e.

V m = V(X)/ ν(x),

where V m is the molar volume of gas - a constant value for any gas under given conditions; V(X) – volume of gas X; ν(x) – amount of gas substance X. Molar volume of gases under normal conditions ( normal pressure pH = 101,325 Pa ≈ 101.3 kPa and temperature Tn = 273.15 K ≈ 273 K) is V m = 22.4 l/mol.

In calculations involving gases, it is often necessary to switch from these conditions to normal ones or vice versa. In this case, it is convenient to use the formula following from the combined gas law of Boyle-Mariotte and Gay-Lussac:

──── = ─── (3)

Where p is pressure; V – volume; T - temperature in Kelvin scale; the index “n” indicates normal conditions.

The composition of gas mixtures is often expressed using the volume fraction - the ratio of the volume of a given component to the total volume of the system, i.e.

where φ(X) is the volume fraction of component X; V(X) – volume of component X; V is the volume of the system. Volume fraction is a dimensionless quantity; it is expressed in fractions of a unit or as a percentage.

7. Which one volume will take at a temperature of 20 o C and a pressure of 250 kPa ammonia weighing 51 g?

Given: m(NH 3)=51 g; p=250 kPa; t=20 o C.

Find: V(NH 3) =?

Solution: determine the amount of ammonia substance:

ν(NH 3) = m(NH 3)/ M(NH 3) = 51/17 = 3 mol.

The volume of ammonia under normal conditions is:

V(NH 3) = V m ν(NH 3) = 22.4 3 = 67.2 l.

Using formula (3), we reduce the volume of ammonia to these conditions [temperature T = (273 +20) K = 293 K]:

p n TV n (NH 3) 101.3 293 67.2

V(NH 3) =──────── = ───────── = 29.2 l.

8. Define volume, which will be occupied under normal conditions by a gas mixture containing hydrogen, weighing 1.4 g, and nitrogen, weighing 5.6 g.

Given: m(N 2)=5.6 g; m(H 2)=1.4; Well.

Find: V(mixtures)=?

Solution: find the amounts of hydrogen and nitrogen substances:

ν(N 2) = m(N 2)/ M(N 2) = 5.6/28 = 0.2 mol

ν(H 2) = m(H 2)/ M(H 2) = 1.4/ 2 = 0.7 mol

Since under normal conditions these gases do not interact with each other, the volume gas mixture will be equal to the sum of the volumes of gases, i.e.

V(mixtures)=V(N 2) + V(H 2)=V m ν(N 2) + V m ν(H 2) = 22.4 0.2 + 22.4 0.7 = 20.16 l.

Calculations using chemical equations

Calculations using chemical equations (stoichiometric calculations) are based on the law of conservation of mass of substances. However, in real chemical processes, due to incomplete reaction and various losses of substances, the mass of the resulting products is often less than that which should be formed in accordance with the law of conservation of mass of substances. The yield of the reaction product (or mass fraction of yield) is the ratio, expressed as a percentage, of the mass of the actually obtained product to its mass, which should be formed in accordance with the theoretical calculation, i.e.

η = /m(X) (4)

Where η is the product yield, %; m p (X) - mass of product X obtained in real process; m(X) – calculated mass of substance X.

In those tasks where the product yield is not specified, it is assumed that it is quantitative (theoretical), i.e. η=100%.

9. What mass of phosphorus needs to be burned for getting phosphorus (V) oxide weighing 7.1 g?

Given: m(P 2 O 5) = 7.1 g.

Find: m(P) =?

Solution: we write down the equation for the combustion reaction of phosphorus and arrange the stoichiometric coefficients.

4P+ 5O 2 = 2P 2 O 5

Determine the amount of substance P 2 O 5 resulting in the reaction.

ν(P 2 O 5) = m(P 2 O 5)/ M(P 2 O 5) = 7.1/142 = 0.05 mol.

From the reaction equation it follows that ν(P 2 O 5) = 2 ν(P), therefore, the amount of phosphorus required in the reaction is equal to:

ν(P 2 O 5)= 2 ν(P) = 2 0.05= 0.1 mol.

From here we find the mass of phosphorus:

m(P) = ν(P) M(P) = 0.1 31 = 3.1 g.

10. Magnesium weighing 6 g and zinc weighing 6.5 g were dissolved in excess hydrochloric acid. What volume hydrogen, measured under standard conditions, will stand out wherein?

Given: m(Mg)=6 g; m(Zn)=6.5 g; Well.

Find: V(H 2) =?

Solution: we write down the reaction equations for the interaction of magnesium and zinc with hydrochloric acid and arrange the stoichiometric coefficients.

Zn + 2 HCl = ZnCl 2 + H 2

Mg + 2 HCl = MgCl 2 + H 2

We determine the amounts of magnesium and zinc substances that reacted with hydrochloric acid.

ν(Mg) = m(Mg)/ М(Mg) = 6/24 = 0.25 mol

ν(Zn) = m(Zn)/ M(Zn) = 6.5/65 = 0.1 mol.

From the reaction equations it follows that the amounts of metal and hydrogen substances are equal, i.e. ν(Mg) = ν(H 2); ν(Zn) = ν(H 2), we determine the amount of hydrogen resulting from two reactions:

ν(H 2) = ν(Mg) + ν(Zn) = 0.25 + 0.1 = 0.35 mol.

We calculate the volume of hydrogen released as a result of the reaction:

V(H 2) = V m ν(H 2) = 22.4 0.35 = 7.84 l.

11. When a volume of 2.8 liters of hydrogen sulfide (normal conditions) was passed through an excess solution of copper (II) sulfate, a precipitate weighing 11.4 g was formed. Determine the exit reaction product.

Given: V(H 2 S)=2.8 l; m(sediment)= 11.4 g; Well.

Find: η =?

Solution: we write down the equation for the reaction between hydrogen sulfide and copper (II) sulfate.

H 2 S + CuSO 4 = CuS ↓+ H 2 SO 4

We determine the amount of hydrogen sulfide involved in the reaction.

ν(H 2 S) = V(H 2 S) / V m = 2.8/22.4 = 0.125 mol.

From the reaction equation it follows that ν(H 2 S) = ν(СuS) = 0.125 mol. This means we can find the theoretical mass of CuS.

m(СuS) = ν(СuS) М(СuS) = 0.125 96 = 12 g.

Now we determine the product yield using formula (4):

η = /m(X)= 11.4 100/ 12 = 95%.

12. Which one weight ammonium chloride is formed by the interaction of hydrogen chloride weighing 7.3 g with ammonia weighing 5.1 g? Which gas will remain in excess? Determine the mass of the excess.

Given: m(HCl)=7.3 g; m(NH 3)=5.1 g.

Find: m(NH 4 Cl) =? m(excess) =?

Solution: write down the reaction equation.

HCl + NH 3 = NH 4 Cl

This task is about “excess” and “deficiency”. We calculate the amounts of hydrogen chloride and ammonia and determine which gas is in excess.

ν(HCl) = m(HCl)/ M(HCl) = 7.3/36.5 = 0.2 mol;

ν(NH 3) = m(NH 3)/ M(NH 3) = 5.1/ 17 = 0.3 mol.

Ammonia is in excess, so we calculate based on the deficiency, i.e. for hydrogen chloride. From the reaction equation it follows that ν(HCl) = ν(NH 4 Cl) = 0.2 mol. Determine the mass of ammonium chloride.

m(NH 4 Cl) = ν(NH 4 Cl) М(NH 4 Cl) = 0.2 53.5 = 10.7 g.

We have determined that ammonia is in excess (in terms of the amount of substance, the excess is 0.1 mol). Let's calculate the mass of excess ammonia.

m(NH 3) = ν(NH 3) M(NH 3) = 0.1 17 = 1.7 g.

13. Technical calcium carbide weighing 20 g was treated with excess water, obtaining acetylene, which, when passed through excess bromine water, formed 1,1,2,2-tetrabromoethane weighing 86.5 g. Determine mass fraction CaC 2 in technical carbide.

Given: m = 20 g; m(C 2 H 2 Br 4) = 86.5 g.

Find: ω(CaC 2) =?

Solution: we write down the equations for the interaction of calcium carbide with water and acetylene with bromine water and arrange the stoichiometric coefficients.

CaC 2 +2 H 2 O = Ca(OH) 2 + C 2 H 2

C 2 H 2 +2 Br 2 = C 2 H 2 Br 4

Find the amount of tetrabromoethane substance.

ν(C 2 H 2 Br 4) = m(C 2 H 2 Br 4)/ M(C 2 H 2 Br 4) = 86.5/ 346 = 0.25 mol.

From the reaction equations it follows that ν(C 2 H 2 Br 4) = ν(C 2 H 2) = ν(CaC 2) = 0.25 mol. From here we can find the mass of pure calcium carbide (without impurities).

m(CaC 2) = ν(CaC 2) M(CaC 2) = 0.25 64 = 16 g.

We determine the mass fraction of CaC 2 in technical carbide.

ω(CaC 2) =m(CaC 2)/m = 16/20 = 0.8 = 80%.

Solutions. Mass fraction of solution component

14. Sulfur weighing 1.8 g was dissolved in benzene with a volume of 170 ml. The density of benzene is 0.88 g/ml. Define mass fraction sulfur in solution.

Given: V(C 6 H 6) = 170 ml; m(S) = 1.8 g; ρ(C 6 C 6) = 0.88 g/ml.

Find: ω(S) =?

Solution: to find the mass fraction of sulfur in a solution, it is necessary to calculate the mass of the solution. Determine the mass of benzene.

m(C 6 C 6) = ρ(C 6 C 6) V(C 6 H 6) = 0.88 170 = 149.6 g.

Find the total mass of the solution.

m(solution) = m(C 6 C 6) + m(S) = 149.6 + 1.8 = 151.4 g.

Let's calculate the mass fraction of sulfur.

ω(S) =m(S)/m=1.8 /151.4 = 0.0119 = 1.19%.

15. Iron sulfate FeSO 4 7H 2 O weighing 3.5 g was dissolved in water weighing 40 g. Determine mass fraction of iron (II) sulfate in the resulting solution.

Given: m(H 2 O)=40 g; m(FeSO 4 7H 2 O) = 3.5 g.

Find: ω(FeSO 4) =?

Solution: find the mass of FeSO 4 contained in FeSO 4 7H 2 O. To do this, calculate the amount of the substance FeSO 4 7H 2 O.

ν(FeSO 4 7H 2 O)=m(FeSO 4 7H 2 O)/M(FeSO 4 7H 2 O)=3.5/278=0.0125 mol

From the formula of iron sulfate it follows that ν(FeSO 4) = ν(FeSO 4 7H 2 O) = 0.0125 mol. Let's calculate the mass of FeSO 4:

m(FeSO 4) = ν(FeSO 4) M(FeSO 4) = 0.0125 152 = 1.91 g.

Considering that the mass of the solution consists of the mass of iron sulfate (3.5 g) and the mass of water (40 g), we calculate the mass fraction of ferrous sulfate in the solution.

ω(FeSO 4) =m(FeSO 4)/m=1.91 /43.5 = 0.044 =4.4%.

Problems to solve independently

50 g of methyl iodide in hexane were exposed to metallic sodium, and 1.12 liters of gas were released, measured under normal conditions. Determine the mass fraction of methyl iodide in the solution. Answer: 28,4%.
Some alcohol was oxidized to form a monobasic carboxylic acid. When 13.2 g of this acid was burned, carbon dioxide was obtained, the complete neutralization of which required 192 ml of KOH solution with a mass fraction of 28%. The density of the KOH solution is 1.25 g/ml. Determine the formula of alcohol. Answer: butanol.
The gas obtained by reacting 9.52 g of copper with 50 ml of an 81% nitric acid solution with a density of 1.45 g/ml was passed through 150 ml of a 20% NaOH solution with a density of 1.22 g/ml. Determine the mass fractions of dissolved substances. Answer: 12.5% NaOH; 6.48% NaNO 3 ; 5.26% NaNO2.
Determine the volume of gases released during the explosion of 10 g of nitroglycerin. Answer: 7.15 l.
Sample organic matter weighing 4.3 g was burned in oxygen. The reaction products are carbon monoxide (IV) with a volume of 6.72 l (normal conditions) and water with a mass of 6.3 g. Vapor density starting material for hydrogen it is 43. Determine the formula of the substance. Answer: C 6 H 14.

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