Algorithms for solving problems in chemistry. How to find the mass fraction of a substance by the formula
Using the example of freezing of a salt solution, you made sure that the presence of foreign substances changes the properties of the substance. In some areas of technology, the use of insufficiently "clean" materials is unacceptable. A particularly pure silicon crystal is used in a computer microcircuit, in nuclear power there are increased requirements for the purification of nuclear fuel, the light signal "goes out" (does not pass through the fiberglass cable), stumbling upon extraneous inclusions.
If the main (main) substance contains extraneous contaminants, this is also a mixture, only in this case all unnecessary and sometimes harmful components are called in one word - impurities. The fewer impurities, the purer the substance.
Sometimes a substance containing impurities is called a technical sample or simply a sample. Therefore, any such sample includes a base material and impurities.
The degree of purity of a substance is usually expressed by the mass fraction of the main component or the mass fraction of impurities.
You are already familiar with mass fractions of different types. Now try to formulate your own definition of what is the mass fraction of impurities in a substance.
Suppose you need to calculate the mass fraction of a basic substance in a sample. Then you can use the formula
or remember that the sum of the mass fractions of the basic substance and impurities is always 1, or 100%:
w (basic in-va) + w (impurities) = 1, or 100%.
It is also true that the mass of the sample consists of the mass of the main substance and the mass of impurities:
m (sample) = m (main. islands) + m (impurities).
Let's take a look at several tasks using the concept of "mass fraction of impurities".
Problem 8... Natural native sulfur contains 8% of impurities. What mass of pure sulfur is contained in 2 tons of a natural sample?
Problem 9... V Food Industry you can use citric acid containing no more than 1% of impurities. The analytical laboratory found that 2.345 g of the product contains 2.312 g of acid. Can the product be used for food purposes?
2. Let's calculate the mass fraction of impurities in the sample: w (impurities) = 1 - w (acids) = 1 - 0.986 = 0.014, or 1.4%.
Answer. This sample citric acid cannot be used in the food industry.
Questions and tasks
- What is called the mass fraction of impurities? What does this value show?
- The industry uses substances labeled "h", which means "pure substance". The content of impurities in them can be, for example, 0.01%. Find the mass of impurities in a 120 g sample of soot labeled "h".
- Mass fraction of impurities in limestone is 5%. Calculate the mass of the basic substance (calcium carbonate) contained in 300 kg of natural limestone.
- When cleaning copper sulfate obtained 150 mg of impurities, which amounted to 2% of the mass of the sample. Determine the mass of the technical copper sulfate that was purified.
- For the manufacture of semiconductor batteries, ultrapure silicon is used. The mass fraction of impurities in it should not exceed 0.0000000001%. Is silicon suitable for these purposes, 30 kg of which contains 0.03 mg of impurities?
The decision on the need to keep such a notebook did not come immediately, but gradually, with the accumulation of work experience.
In the beginning it was a place at the end workbook- several pages to record the most important definitions. Then the most important tables were placed there. Then the realization came that most students, in order to learn how to solve problems, need strict algorithmic prescriptions, which they, first of all, must understand and remember.
It was then that the decision came to maintain, in addition to the workbook, one more obligatory notebook in chemistry - a chemical dictionary. Unlike workbooks, of which there may even be two within one academic year, a dictionary is a single notebook for the entire chemistry course. It is best if this notebook has 48 sheets and a sturdy cover.
We arrange the material in this notebook as follows: at the beginning - the most important definitions that the guys write out from the textbook or write down under the teacher's dictation. For example, in the first lesson in the 8th grade, this is the definition of the subject "chemistry", the concept of "chemical reactions". During the academic year, more than thirty of them accumulate in the 8th grade. For these definitions, I conduct surveys in some lessons. For example, an oral question in a chain, when one student asks a question to another, if he answered correctly, it means that he is already asking the next one; or, when one student is asked questions by other students, if he does not cope with the answer, then they answer themselves. In organic chemistry, these are mainly class definitions organic matter and the main concepts, for example, "homologues", "isomers", etc.
At the end of our reference book, material is presented in the form of tables and diagrams. The very first table “Chemical elements. Chemical signs ”. Then the tables “Valence”, “Acids”, “Indicators”, “Electrochemical series of metal voltages”, “Series of electronegativity”.
I would especially like to dwell on the contents of the table "Correspondence of acids to acid oxides":
| Correspondence of acids to acid oxides | ||||
| Acidic oxide | Acid | |||
| Name | Formula | Name | Formula | Acid residue, valence |
| carbon monoxide (II) | CO 2 | coal | H 2 CO 3 | CO 3 (II) |
| sulfur (IV) oxide | SO 2 | sulphurous | H 2 SO 3 | SO 3 (II) |
| sulfur (VI) oxide | SO 3 | sulfuric | H 2 SO 4 | SO 4 (II) |
| silicon oxide (IV) | SiO 2 | silicon | H 2 SiO 3 | SiO 3 (II) |
| nitric oxide (V) | N 2 O 5 | nitrogen | HNO 3 | NO 3 (I) |
| phosphorus (V) oxide | P 2 O 5 | phosphoric | H 3 PO 4 | PO 4 (III) |
Without understanding and memorizing this table, it is difficult for pupils of the 8th grade to compose the equations of reactions acid oxides with alkalis.
When studying the theory of electrolytic dissociation, we write down the diagrams and rules at the end of the notebook.
Rules for drawing up ionic equations:
1. In the form of ions, write down the formulas of strong electrolytes, soluble in water.
2. In molecular form, write down the formulas of simple substances, oxides, weak electrolytes and all insoluble substances.
3. Formulas of poorly soluble substances on the left side of the equation are written in ionic form, on the right - in molecular form.
When studying organic chemistry we write in the dictionary summarizing tables for hydrocarbons, classes of oxygen and nitrogen-containing substances, schemes for genetic relationships.
| Physical quantities | |||
| Designation | Name | Units | Formulas |
| amount of substance | mole | = N / N A; = m / M; V / V m (for gases) |
|
| N A | Avogadro's constant | molecules, atoms and other particles | N A = 6.02 10 23 |
| N | number of particles | molecules, atoms and other particles |
N = N A |
| M | molar mass | g / mol, kg / kmol | M = m /; / M / = M r |
| m | weight | g, kg | m = M; m = V |
| V m | molar gas volume | l / mol, m 3 / kmol | Vm = 22.4 l / mol = 22.4 m 3 / kmol |
| V | volume | l, m 3 | V = V m (for gases); |
| density | g / ml; | = m / V; M / V m (for gases) |
|
For 25 - summer period teaching chemistry at school, I had to work on different programs and textbooks. At the same time, it was always surprising that practically no textbook teaches how to solve problems. At the beginning of the study of chemistry, in order to systematize and consolidate knowledge in the dictionary, the students and I compile a table "Physical quantities" with new values:
When teaching students how to solve computational problems, I attach great importance to algorithms. I believe that strict sequencing guidelines allow the weak learner to understand how to solve certain types of problems. For strong students, this is an opportunity to reach the creative level of their further chemical education and self-education, since first you need to confidently master a relatively small number of standard techniques. On the basis of this, the ability to correctly apply them at different stages of solving more complex problems will be developed. Therefore, algorithms for solving computational problems have been compiled by me for all types of problems in the school course and for optional classes.
I will give examples of some of them.
Algorithm for solving problems using chemical equations.
1. Write down briefly the condition of the problem and make a chemical equation.
2. Above the formulas in the chemical equation, inscribe the data of the problem, under the formulas, write the number of moles (determined by the coefficient).
3. Find the amount of a substance, the mass or volume of which is given in the problem statement, using the formulas:
M / M; = V / V m (for gases V m = 22.4 l / mol).
Write the resulting number over the formula in the equation.
4. Find the amount of a substance whose mass or volume is unknown. To do this, carry out the reasoning according to the equation: compare the number of moles according to the condition with the number of moles according to the equation. If necessary, make up the proportion.
5. Find the mass or volume by the formulas: m = M; V = V m.
This algorithm is the basis that a student must master so that in the future he can solve problems using equations with various complications.
Excess and deficiency problems.
If in the condition of the problem the quantities, masses or volumes of two reacting substances are known at once, then this is a problem for excess and deficiency.
When solving it:
1. It is necessary to find the quantities of two reactants according to the formulas:
M / M; = V / V m.
2. The resulting numbers are moles to inscribe over the equation. Comparing them with the number of moles according to the equation, make a conclusion about what substance is given in the deficiency.
3. If there is a shortage, make further calculations.
Tasks on the proportion of the yield of the reaction product, practically obtained from the theoretically possible.
According to the reaction equations, theoretical calculations are carried out and theoretical data for the reaction product are found: theor. , m theor. or V theor. ... When carrying out reactions in the laboratory or in industry, losses occur, therefore the obtained practical data are practical. ,
m pract. or V practical. always less than theoretically calculated data. The percentage of the output is designated by the letter (eta) and is calculated by the formulas:
(this) = practical. / theor. = m practical / m theory. = V practical / V theor.
Express it in fractions of one or as a percentage. Three types of tasks can be distinguished:
If in the condition of the problem the data for the initial substance and the proportion of the yield of the reaction product are known, then it is necessary to find practical. , m pract. or V practical. the reaction product.
Solution procedure:
1. Make a calculation using the equation, based on the data for the original substance, find theor. , m theor. or V theor. the reaction product;
2. Find the mass or volume of the reaction product, practically obtained, according to the formulas:
m pract. = m theor. ; V practical. = V theor. ; practical = theor. ...
If in the condition of the problem the data for the initial substance and practical are known. , m pract. or V practical. of the obtained product, while it is necessary to find the fraction of the yield of the reaction product.
Solution procedure:
1. Make a calculation using the equation, based on the data for the original substance, find
Theor. , m theor. or V theor. the reaction product.
2. Find the fraction of the yield of the reaction product by the formulas:
Practice. / theor. = m practical / m theory. = V practical / V theor.
If in the condition of the problem practical. , m pract. or V practical. of the obtained reaction product and the fraction of its yield, while it is necessary to find the data for the starting material.
Solution procedure:
1. Find theor., M theor. or V theor. the reaction product according to the formulas:
Theor. = practical /; m theor. = m practical /; V theor. = V practical /.
2. Calculate the equation based on the theory. , m theor. or V theor. the reaction product and find the data for the starting material.
Of course, we consider these three types of problems gradually, we practice the skills of solving each of them by the example of a number of problems.
Problems on mixtures and impurities.
The pure substance is the one of which there is more in the mixture, the rest is impurities. Legend: the mass of the mixture is m cm., The mass is pure substance- m h.h., the mass of impurities - m approx. , mass fraction of pure substance - ch.w.
The mass fraction of a pure substance is found by the formula: = m h.v. / m see, express it in fractions of one or as a percentage. Let's select 2 types of tasks.
If in the problem statement the mass fraction of pure substance or the mass fraction of impurities is given, then the mass of the mixture is given. The word "technical" also means the presence of a mixture.
Solution procedure:
1. Find the mass of a pure substance by the formula: m h.v. = h.v. m see
If the mass fraction of impurities is given, then first you need to find the mass fraction of the pure substance: = 1 - approx.
2. Based on the mass of the pure substance, make further calculations using the equation.
If the condition of the problem gives the mass of the initial mixture and n, m or V of the reaction product, then you need to find the mass fraction of the pure substance in the original mixture or the mass fraction of impurities in it.
Solution procedure:
1. Make a calculation according to the equation, based on the data for the reaction product, and find n p.h. and m h.v.
2. Find the mass fraction of a pure substance in the mixture by the formula: h.v. = m h.v. / m see and mass fraction of impurities: approx. = 1 - h. In
The law of volumetric relations of gases.
The volumes of gases are related in the same way as their quantities of substances:
V 1 / V 2 = 1/2
This law is used when solving problems according to equations in which the volume of gas is given and you need to find the volume of another gas.
Volume fraction of gas in the mixture.
Vg / Vcm, where (phi) is the gas volume fraction.
Vg - gas volume, Vcm - gas mixture volume.
If the volume fraction of gas and the volume of the mixture are given in the condition of the problem, then, first of all, you need to find the volume of gas: Vg = Vcm.
The volume of the gas mixture is found by the formula: Vcm = Vg /.
The volume of air spent on burning a substance is found through the volume of oxygen, found by the equation:
Vair. = V (O 2) / 0.21
Derivation of formulas of organic substances according to general formulas.
Organic substances form homologous series that have general formulas. This allows:
1. Express the relative molecular weight in terms of the number n.
M r (C n H 2n + 2) = 12 n + 1 (2n + 2) = 14n + 2.
2. Equate M r, expressed in terms of n, to the true M r and find n.
3. Make up the reaction equations in general view and make calculations on them.
Derivation of formulas of substances by combustion products.
1. Analyze the composition of combustion products and draw a conclusion about the qualitative composition of the burnt substance: Н 2 О -> Н, СО 2 -> С, SO 2 -> S, P 2 O 5 -> P, Na 2 CO 3 -> Na, C.
The presence of oxygen in the substance requires verification. Designate the indices in the formula by x, y, z. For example, СxНyОz (?).
2. Find the amount of substances in the combustion products by the formulas:
n = m / M and n = V / Vm.
3. Find the number of elements contained in the burnt substance. For example:
n (C) = n (CO 2), n (H) = 2 ћ n (H 2 O), n (Na) = 2 ћ n (Na 2 CO 3), n (C) = n (Na 2 CO 3), etc.
4. If a substance of unknown composition has burned out, it is imperative to check whether it contains oxygen. For example, CxHyOz (?), M (O) = m in – va - (m (C) + m (H)).
b) if the relative density is known: M 1 = D 2 M 2, M = D H2 2, M = D O2 32,
M = D air. 29, M = D N2 28, etc.
Method 1: find the simplest formula substances (see the previous algorithm) and the simplest molar mass. Then compare the true molar mass with the simplest one and increase the indices in the formula by the required number of times.
Method 2: find the indices by the formula n = (e) Mr / Ar (e).
If the mass fraction of one of the elements is unknown, then it must be found. To do this, subtract the mass fraction of another element from 100% or from the unit.
Gradually, in the course of studying chemistry in the chemical dictionary, algorithms for solving problems are accumulated different types... And the student always knows where to find the right formula or the right information to solve the problem.
Many students like keeping such a notebook, they themselves supplement it with various reference materials.
As for the extracurricular activities, the students and I also start a separate notebook to write down algorithms for solving problems that go beyond school curriculum... In the same notebook for each type of problem we write down 1-2 examples, the rest of the problems they solve in another notebook. And, if you think about it, among the thousands of different tasks encountered on the chemistry exam in all universities, tasks of 25-30 different types can be distinguished. Of course, there are many variations among them.
In the development of algorithms for solving problems in optional classes, the manual by A.A. Kushnareva. (Learning to solve problems in chemistry, - M., School - press, 1996).
The ability to solve problems in chemistry is the main criterion for the creative assimilation of a subject. It is through solving problems of various levels of complexity that a chemistry course can be effectively mastered.
If a student has a clear idea of all possible types of problems, solved a large number of problems of each type, then he is able to cope with passing the exam in chemistry in the form of the Unified State Exam and upon entering universities.
The space around us is filled with different physical bodies, which are composed of different substances with different weights. School courses in chemistry and physics, introducing the concept and method of finding the mass of a substance, were listened to and safely forgotten by everyone who attended school. But meanwhile, theoretical knowledge acquired sometime may be needed at the most unexpected moment.
Calculation of the mass of a substance using the specific gravity of a substance. Example - there is a 200 liter barrel. You need to fill the barrel with any liquid, say a light beer. How do you find the mass of a full barrel? Using the formula for the density of a substance p = m / V, where p is the specific density of the substance, m is the mass, V is the occupied volume, it is very simple to find the mass of a full barrel:- Volume measures - cubic centimeters, meters. That is, a 200 liter barrel has a volume of 2 m³.
- A measure of specific gravity is found using tables and is constant for each substance. The density is measured in kg / m³, g / cm³, t / m³. The density of light beer and other alcoholic beverages can be viewed on the website. It is 1025.0 kg / m³.
- From the density formula p = m / V => m = p * V: m = 1025.0 kg / m³ * 2 m³ = 2050 kg.
A barrel with a volume of 200 liters, completely filled with light beer, will have a mass of 2050 kg.
Finding the mass of a substance using molar mass. M (x) = m (x) / v (x) is the ratio of the mass of a substance to its amount, where M (x) is the molar mass of X, m (x) is the mass of X, v (x) is the amount of substance X If only 1 known parameter is prescribed in the problem statement - the molar mass of a given substance, then finding the mass of this substance will not be difficult. For example, you need to find the mass of sodium iodide NaI with an amount of 0.6 mol.- Molar mass is calculated in the SI unified system of measurements and is measured in kg / mol, g / mol. The molar mass of sodium iodide is the sum of the molar masses of each element: M (NaI) = M (Na) + M (I). The value of the molar mass of each element can be calculated from the table, or you can use the online calculator on the website: M (NaI) = M (Na) + M (I) = 23 + 127 = 150 (g / mol).
- From the general formula M (NaI) = m (NaI) / v (NaI) => m (NaI) = v (NaI) * M (NaI) = 0.6 mol * 150 g / mol = 90 grams.
The mass of sodium iodide (NaI) with a mass fraction of a substance of 0.6 mol is 90 grams.
- Dilute the solution with water. The mass of the dissolved substance X does not change m (X) = m '(X). The mass of the solution increases by the mass of added water m '(p) = m (p) + m (H 2 O).
- Evaporation of water from solution. The mass of the solute X does not change m (X) = m '(X). The mass of the solution decreases by the mass of evaporated water m '(p) = m (p) -m (H 2 O).
- Draining two solutions. The masses of the solutions, as well as the masses of the solute X, when mixed, add up: m ’’ (X) = m (X) + m ’(X). m '' (p) = m (p) + m '(p).
- Loss of crystals. The masses of the solute X and the solution decrease by the mass of the precipitated crystals: m '(X) = m (X) -m (sediment), m' (p) = m (p) -m (sediment).
Options for finding the mass of matter - a useful course schooling, but methods that are quite applied in practice. Everyone can easily find the mass of the necessary substance by applying the above formulas and using the proposed tables. To facilitate the task, write down all the reactions, their coefficients.
Students often have difficulty solving problems. The design algorithm and problem solving will help them with this. Algorithms for solving problems by topics are considered here:
- DETERMINATION OF THE MASS SHARE OF THE PRODUCT OF THE REACTION FROM THE THEORETICALLY POSSIBLE.
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"Algorithms for solving problems in chemistry"
Algorithm
solving problems on impurities.
Determination of the mass (volume) of a substance by known mass another substance containing a certain proportion of impurities.
Remember: 1. The peculiarity of this type of problems is that first it is necessary to calculate the mass of a pure substance in a mixture.
2. In the condition of the problem, ore, a technical sample of a substance, or a solution can act as a mixture.
Scheme for solving the problem :
Analysis and short calculations for
Which record is the definition of the equation is the answer.
Conditions for the problem of the mass of a pure reaction
substances
Procedure for solving the problem:
Determine the mass of the pure substance by the formula: m in-va = m mixes * ω in-va.
Write the equation for the reaction.
Find the quantities of substances given in the problem by the equation and by the condition.
Produce necessary calculations and write down the answer.
Sample solution:
Calculate the volume of hydrogen evolved by reacting with hydrochloric acid 325 g of zinc containing 20% impurities.
D a n o: Decision:
m technical (Zn) = 325 g 1) m technical (Zn) = 325 g ω (Zn) = 100% -20% = 80% (0.8);
ω approx. = 20% (0,2) ω approx. = 20% (0.2) m (Zn) = 325 * 0.8 = 260 g
V (H 2) =? n (Zn) = 260g: 65 g / mol = 4 mol.
by condition: 4 mol X mol
2) Zn + 2HCl = ZnCl 2 + H 2
by equation: 1 mol 1 mol
V (H 2) = Vm * n (H 2); V (H 2) = 22.4 l / mol * 4 mol = 89.6 l.
Answer: V (H 2) = 89.6 liters.
Algorithm
solving problems on impurities.
Determination of the mass fraction of impurities (or the mass fraction of a pure substance in a mixture) by mass (volume) of reaction products.
Remember : 1) first, actions are performed according to the reaction equation;
2) to determine the mass fraction of impurities, we use the formula:
ω approx. = m approx. / m mixture.
Scheme for solving the problem :
tasks mass calculations definition definition
Substances according to the equation of mass - mass fraction answer.
No impurities impurities
Procedure for solving the problem :
Read the problem, write down a short condition.
Write the equation for the chemical reaction.
Calculate the mass of the pure substance required for the reaction according to the reaction equation .
Calculate the mass of impurities in the sample by condition.
Calculate the mass fraction of impurities using the formula: ω approx. = m approx. / m mixture.
Sample solution:
Determine the mass fraction of impurities in a technical sample of calcium carbide if 56 liters of acetylene were obtained from 200 g of it.
Given: Solution: let xg be the mass of pure substance CaC 2 .
m technical (CaC 2) = 200g 1) by condition: xg 56l
V (C 2 H 2) = 56 l CaC 2 + 2H 2 O = WITH 2 N 2 + Ca (OH) 2
ω approx =? 1 mol 1 mol
М = 64g / mol V m = 22.4l / mol
according to the equation: m = 64 g V = 22.4 l,
then x g / 64 g = 56 l / 22.4 l; x = 160 g.
2) determine the mass of impurities in the sample:
m approx. = 200 - 160 = 40 g.
3) determine the mass fraction of impurities:
ω approx = 40 g / 200 g = 0.2 (or 20%).
Answer: ω approx = 20%.
ALGORITHM FOR SOLVING AND FORMATTING THE PROBLEMS FOR DETERMINING THE MASS RATIO OF THE OUTPUT OF THE REACTION PRODUCT FROM THEORETICALLY POSSIBLE.
| 1. Read the condition carefully | When passing 11.2 liters of ammonia through the solution nitric acid, received 15 g of ammonium nitrate. How much% is it from theoretically possible exit. |
| 2. Write down "Given:" (the mass of the reaction product given by condition is its practical | V (NH 3) = 11.2L m practical out (NH 4 NO 3 ) = 15g_ η (NH 4 NO 3) -? |
| 3. Write down the formula for the mass fraction the yield of the reaction product from theoretically possible | |
| 4. Write down the reaction equation. Arrange the coefficients. Underline the formulas of those substances with which you will solve the problem. | NH 3 + HNO 3 → NH 4 NO 3 |
| 5. Above the formula of the original substance fill in the mass (V), data by condition, and above the formula put the reaction product - NS, So you will find mass (V) him theoretical output | NH 3 + HNO 3 → NH 4 NO 3 |
| 6. Under the formulas underlined substances: fill in their number moles; define and put down molar masses (volumes) of these substances, not forgetting to multiply them by the number of moles. | NH 3 + HNO 3 → NH 4 NO 3 1 mol 1 mol × 22.4 l / mol × 80 g / mol |
| 7. Proportion, find the value of X - this will be mass (volume) of theoretical the yield of the reaction product. ... | NH 3 + HNO 3 → NH 4 NO 3 1 mol 1 mol × 22.4 l / mol × 80 g / mol 22.4L = 80g = X = = 40g - m theoretical out |
| 7. Substitute the practical and theoretical yield of the reaction product into the mass formula (V) and calculate the mass fraction of the yield of the reaction product | ||
| 8. Write down your answer. | Answer: η (NH 4 NO 3) = 37.5% |
|
Sample task design:
Given: Solution:
V (NH 3) = 11.2L 1) m (V) theoretical exit -?
m practical out (NH 4 NO 3 ) = 15g 11,2L X
η (NH 4 NO 3) -? NH 3 + HNO 3 → NH 4 NO 3
1 mol 1 mol
× 22.4 l / mol × 80 g / mol
X = = 40g - m theoretical out
2) η (NH 4 NO 3) -?
η (NH 4 NO 3) = 100% = 100% = 37.5%
Answer: η (NH 4 NO 3) = 37.5%
Decide for yourself:
By passing 170 g of ammonia through a hydrochloric acid solution, 500 g of ammonium chloride was obtained. What is the% of the theoretically possible yield
Algorithm for solving problems
Calculation by chemical equations
Calculations by chemical equations of reactions, if one substance is taken in excess
| Actions | ||
| We write down the condition of the problem ("given") | m (K 2 CO 3) = 27.6g m (HNO 3) = 315 g |
|
| We compose the reaction equation (we correctly write down all the formulas of substances, according to the valences, we arrange the coefficients) | m (K 2 CO 3) = 27.6g m (HNO 3) = 315 g |
|
| Above the reaction equation, we write out the data from the condition of the problem | m (K 2 CO 3) = 27.6g m (HNO 3) = 315 g K 2 CO 3 +2 HNO 3 → 2 KNO 3 + CO 2 + H 2 O |
|
| Under the equation, we indicate the number of moles according to the reaction equation (the number of moles is determined by the coefficient, only for substances that are indicated in the condition) | m (K 2 CO 3) = 27.6g m (HNO 3) = 315 g K 2 CO 3 +2 HNO 3 → 2 KNO 3 + CO 2 + H 2 O 1 mol 2 mol 1 mol |
|
| Convert mass (volume) to moles | m (K 2 CO 3) = 27.6g m (HNO 3) = 315 g K 2 CO 3 +2 HNO 3 → 2 KNO 3 + CO 2 + H 2 O 1 mol 2 mol 1 mol Мr (HNO 3) = 1 + 14 + 16 ∙ 3 = 63 n (HNO 3) = 315/63 = 5 mol |
|
| We determine which substance is in excess and which is in short supply. To do this, select the smallest value of the resulting moles and make up the proportion. We solve the problem by lack. | m (K 2 CO 3) = 27.6g m (HNO 3) = 315 g 0.2 mol 5 mol K 2 CO 3 +2 HNO 3 → 2 KNO 3 + CO 2 + H 2 O 1 mol 2 mol 1 mol n = m / M Мr (K 2 CO 3) = 39 ∙ 2 + 12 + 16 ∙ 3 = 138 Мr (HNO 3) = 1 + 14 + 16 ∙ 3 = 63 n (K 2 CO 3) = 27.6 / 138 = 0.2 mol n (HNO 3) = 315/63 = 5 mol 0.2 x x = 0.2 ∙ 2/1 = 0.4 mol One can argue as follows: according to the equation for 1 mol of K 2 CO 3, 2 mol of HNO 3 is needed (twice as much), which means that for 0.2 mol of K 2 CO 3, 0.4 mol of HNO 3 is enough, and according to the condition of the problem, 5 mol (315g ). |
|
| We make up the proportion (write out through a fraction the upper and lower values of the moles for). We decide the proportion. | ----------- ------- 1 mol 1 mol X = 0.2 mol (CO 2) |
|
| Calculate the volume of carbon monoxide (IV) | V (CO 2) = 0.2 mol ∙ 22.4 mol / l = 4.48L |
