CALCULATION OF THE YIELD OF THE REACTION PRODUCT IN PERCENTAGE OF THE THEORETICALLY POSSIBLE, IF THE MASSES OF THE INITIAL SUBSTANCE AND THE REACTION PRODUCT ARE KNOWN

Objective 1. Carbon dioxide was passed through lime water containing 3.7 g of calcium hydroxide. The precipitate that formed was filtered off, dried, and weighed. Its mass turned out to be equal to 4.75 g. Calculate the yield of the reaction product (in percent) of the theoretically possible.

Given:

Method I.

Let us determine the amount of substances given in the condition of the problem:
v \u003d m / M \u003d 3.7 g / 74 g / mol \u003d 0.05 mol;
v \u003d 0.05 mol
v (CaCO 3 ) \u003d m (CaCO 3 ) / M (CaCO 3 ) \u003d 4.75 g / 100 g / mol \u003d 0.0475 mol;
v (CaCO 3 ) \u003d 0.0475 mol

Let's write the equation chemical reaction:

Ca (OH) 2

It follows from the chemical reaction equation that from 1 mol of Ca (OH) 2 1 mol of CaCO is formed 3 , which means that from 0.05 mol of Ca (OH) 2 theoretically, the same should be obtained, that is, 0.05 mol of CaCO 3 ... In practice, 0.0475 mol of CaCO was obtained 3 , which will be:
w out. (CaCO 3 ) \u003d 0.0475 mol * 100 % / 0.05 mol \u003d 95%
w out. (CaCO 3 ) = 95 %

Method II.

We take into account the mass of the initial substance (calcium hydroxide) and the equation of the chemical reaction:

Ca (OH) 2

Let's calculate according to the reaction equation how much calcium carbonate is theoretically formed.

From 74 g Ca (OH) 2

Hence x \u003d 3.7 g* 100 g/ 74 g \u003d 5 g, m (CaCO 3 ) \u003d 5 g

This means that from the data on the condition of the problem, 3.7 g of calcium hydroxide, theoretically (from calculations), it would be possible to obtain 5 g of calcium carbonate, and in practice, only 4.75 g of the reaction product was obtained. From these data, we determine the yield of calcium carbonate (in%) of the theoretically possible:

5 g CaCO 3 make up 100% yield
4.75 g CaCO 3 make up x%

x \u003d 4.75 mol* 100 % / 5 g \u003d 95%;
w out. (CaCO 3 ) = 95 %

Answer: The yield of calcium carbonate is 95% of the theoretically possible.

Objective 2. The interaction of magnesium weighing 36 g with an excess of chlorine yielded 128.25 g of magnesium chloride. Determine the yield of the reaction product as a percentage of the theoretically possible.

Given: Consider two ways to solve this problem: using the quantity amount of substance and mass of matter.

Method I.

From the data on the condition of the problem of the values \u200b\u200bof the masses of magnesium and magnesium chloride, we calculate the values \u200b\u200bof the amount of these substances:
v (Mg) \u003d m (Mg) / M (Mg) \u003d 36 g / 24 g / mol \u003d 1.5 mol; v (Mg) \u003d 1.5 mol
v (MgCl 2 ) \u003d m (MgCl 2 )/ M (MgCl 2 ) \u003d 128.25 g / 95 g / mol \u003d 1.35 mol;
v (MgCl 2 ) \u003d 1.35 mol

Let's compose the equation of the chemical reaction:

Mg

Let's use the equation of a chemical reaction. From this equation it follows that from 1 mole of magnesium you can get 1 mole of magnesium chloride, which means that from the data 1.5 mole of magnesium you can theoretically get the same amount, that is, 1.5 mole of magnesium chloride. Practically only 1.35 mol was obtained. Therefore, the yield of magnesium chloride (in%) of the theoretically possible will be:

1.5 mol MgCl 2

x \u003d 1.35 mol * 100%/ 1.5 mol \u003d 90%, i.e. w out. (MgCl 2 ) = 90%

Method II.

Consider the equation of a chemical reaction:

Mg

First of all, according to the chemical reaction equation, we determine how many grams of magnesium chloride can be obtained from the data on the condition of the problem of 36 g of magnesium.

From 24 g Mg 2

Hence x \u003d 36 g * 95 g/ 24 g \u003d 142.5 g; m (MgCl 2 ) \u003d 142.5 g

This means that 142.5 g of magnesium chloride could be obtained from a given amount of magnesium (theoretical yield 100%). And received only 128.25 g of magnesium chloride (practical yield).
Consider now how many percent is the practical output of the theoretically possible:

142.5 g MgCl 2

x \u003d 128.25 g * 100% / 142.5 g \u003d 90%, that is w out. (MgCl 2 ) = 90%

Answer: the yield of magnesium chloride is 90% of the theoretically possible.

Objective 3. Metallic potassium weighing 3.9 g was placed in distilled water with a volume of 50 ml. As a result of the reaction, 53.8 g of a potassium hydroxide solution with a mass fraction of the substance equal to 10% were obtained. Calculate the yield of caustic potassium (as a percentage) of the theoretically possible.

Given:

2K

Based on this equation of chemical reaction, we will make calculations.
First, let us determine the mass of caustic potassium, which theoretically could be obtained from the potassium mass given by the condition of the problem.

From 78 g K

Hence: x \u003d 3.9 g * 112 g / 78 g \u003d 5.6 g m (KOH) \u003d 5.6 g

From this formula, we express m in-islands:
m in-va \u003d m solution * w in / 100%

Let us determine the mass of caustic potash, which is in 53.8 g of its 10% solution:
m (KOH) \u003d m solution * w (KOH) / 100% \u003d 53.8 g * 10% / 100% \u003d 5.38 g
m (KOH) \u003d 5.38 g

Finally, we calculate the yield of caustic potash as a percentage of the theoretically possible:
w out. (KOH) \u003d 5.38 g / 5.6 g * 100% = 96%
w out. (KOH) \u003d 96%

Answer: The yield of caustic potash is 96% of the theoretically possible.

Atoms and molecules are the smallest particles of matter, therefore, as a unit of measurement, you can choose the mass of one of the atoms and express the masses of other atoms in relation to the selected one. So what is molar mass, and what is its dimension?

What is molar mass?

The founder of the theory of atomic masses was the scientist Dalton, who compiled a table of atomic masses and took the mass of a hydrogen atom as a unit.

Molar mass is the mass of one mole of a substance. A mole, in turn, is the amount of a substance that contains a certain amount of the smallest particles that participate in chemical processes. The number of molecules contained in one mole is called Avogadro's number. This value is constant and does not change.

Figure: 1. Formula of Avogadro's number.

Thus, the molar mass of a substance is the mass of one mole, which contains 6.02 * 10 ^ 23 elementary particles.

Avogadro's number got its name in honor of the Italian scientist Amedeo Avagadro, who proved that the number of molecules in the same volume of gases is always the same

Molar mass in the International System of SI is measured in kg / mol, although this value is usually expressed in grams / mol. This value is denoted english letter M, and the formula for molar mass is:

where m is the mass of a substance, and v is the amount of a substance.

Figure: 2. Calculation of molar mass.

How to find the molar mass of a substance?

DI Mendeleev's table will help to calculate the molar mass of this or that substance. Take any substance, for example, sulfuric acid. Its formula looks like this: H 2 SO 4. Now let's turn to the table and see what the atomic mass of each of the elements that make up the acid is. Sulfuric acid consists of three elements - hydrogen, sulfur, oxygen. The atomic mass of these elements is 1, 32, 16, respectively.

It turns out that the total molecular weight is 98 atomic mass units (1 * 2 + 32 + 16 * 4). Thus, we found out that one mole of sulfuric acid weighs 98 grams.

The molar mass of a substance is numerically equal to the relative molecular mass if the structural units of the substance are molecules. The molar mass of a substance can also be equal to the relative atomic mass if the structural units of the substance are atoms.

Until 1961, an oxygen atom was taken as an atomic mass unit, but not a whole atom, but its 1/16 part. At the same time, chemical and physical unit the masses were not the same. The chemical was 0.03% more than the physical.

At present, a unified measurement system has been adopted in physics and chemistry. As standard ea.m. 1/12 of the mass of the carbon atom is selected.

Figure: 3. Formula of the unit of atomic mass of carbon.

The molar mass of any gas or vapor is very easy to measure. It is enough to use control. One and the same volume of a gaseous substance is equal in quantity to another substance at the same temperature. A known method for measuring the volume of steam is to determine the amount of displaced air. This process is carried out using a lateral arm leading to the measuring device.

The concept of molar mass is very important in chemistry. Its calculation is necessary to create polymer complexes and many other reactions. In pharmaceuticals, the concentration of a given substance in a substance is determined using molar mass. Also, molar mass is important when conducting biochemical studies (exchange process in an element).

In our time, thanks to the development of science, the molecular weights of almost all blood components, including hemoglobin, are known.

Tasks for a practical exit.

1.Calculate the volume of ammonia that can be obtained by heating 20 g of ammonium chloride with an excess of calcium hydroxide if the volume fraction of the ammonia yield is 98%.

2NH 4 Cl + Ca (OH) 2 \u003d 2NH 3 + H 2 O; Mr (NH 4 Cl) \u003d 53.5

NH 4 Cl + 0.5Ca (OH) 2 \u003d NH 3 + 0.5H 2 O

1) Calculate the theoretical output

20 / 53.5 \u003d X / 22.4; X \u003d 8.37L (this is a theoretical yield)

2) Calculate a practical way out

V (practical) \u003d V (theoretical) / product output * 100%

V (practical) \u003d 8.37l * 98% / (divide by) 100% \u003d 8.2l

Answer: 8.2 l N Ns

2. From 320 g of pyrite containing 45% sulfur, 405 g of sulfuric acid was obtained (calculated as anhydrous acid). Calculate the mass fraction of the sulfuric acid yield.

Let's draw up a scheme for the production of sulfuric acid

320g 45% 405g, ή-?

FeS 2 → S → H 2 SO 4

1) Calculate the proportion of sulfur in pyrite

2) Calculate the theoretical yield of sulfuric acid

3) Calculate the product yield as a percentage

H. Calculate the mass of phosphorus required to obtain 200 kg of phosphoric acid, if the mass fraction of the product yield is 90%.

Let's draw up a scheme for the production of phosphoric acid

X 200kg, ή \u003d 90%

P → H 3 PO 4

1) Calculate the mass of the theoretical yield of phosphoric acid

m t \u003d

2) Calculate the mass of phosphorus

Answer: 70, Zkg

4. A young chemist in the classroom decided to get nitric acid by the exchange reaction between potassium nitrate and concentrated sulfuric acid. Calculate the mass of nitric acid, which he received from 20.2 g-potassium nitrate, if the mass fraction of the acid yield was 0.98

5.When ammonium nitrite N H 4 NO 2 is heated, nitrogen and water are formed. Calculate the volume of nitrogen (n.a.) that can be obtained from the decomposition of 6.4 g of ammonium nitrite, if the volume fraction of nitrogen yield is 89%.

6.Calculate the volume of nitric oxide (II) that can be obtained by catalytic oxidation in the laboratory 5.6 liters of ammonia, if the volume fraction of nitrogen oxide (II) output is 90%.

7. Metallic barium is obtained by reduction of its oxide with metallic aluminum with the formation of aluminum oxide and barium. Calculate the mass fraction of the yield of barium if 3.8 kg of barium was obtained from 4.59 kg of barium oxide.

Answer: 92.5%

8. Determine what mass of copper is required for the reaction with an excess of concentrated nitric acid to obtain 2.1 l (n.y) of nitric oxide (IV), if the volume fraction of the yield of nitric oxide (IV) is 94%.

Answer: 3.19

9. What volume of sulfur oxide (IV) should be taken for the oxidation reaction with oxygen to obtain sulfur oxide (V I) with a mass of 20 g. if the product yield is 80% (n.a).?

2SO 2 + O 2 \u003d 2SO 3; V. (SO 2) \u003d 22.4 l; Mr (SO 3 ) =80

1) Calculate the theoretical output

m (theory) \u003d

2) Calculate the mass of SO 2

10. By heating a mixture of calcium oxide weighing 19.6 g with coke weighing 20 g, calcium carbide weighing 16 g was obtained. Determine the yield of calcium carbide if the mass fraction of carbon in the coke is 90%.

Answer: 71.4%

11. An excess of chlorine was passed through a solution weighing 50 g with a mass fraction of sodium iodide of 15%, and iodine weighing 5.6 g was released. Determine the yield of the reaction product from the theoretically possible in%.

Answer: 88.2%.

12. Determine the yield of sodium silicate in% of the theoretical, if by fusing 10 kg of sodium hydroxide with silicon oxide (IV), 12.2 kg of sodium silicate are obtained. Answer 80%

13. From 4 kg of aluminum oxide, 2 kg of aluminum can be smelted. Calculate the mass fraction of the output of aluminum from the theoretically possible.

Answer: 94.3%

14. Calculate the volume of ammonia, which is obtained by heating a mixture of ammonium chloride with a mass of 160.5 g and calcium hydroxide, if the volume fraction of the yield of ammonia from the theoretically possible is 78%.

Answer: 52.4L

15. How much ammonia is required to obtain 8 tons of ammonium nitrate if the product yield is 80% of the theoretically possible?

Answer: 2, ISt

16. What amount of acetaldehyde can be obtained by Kucherov's reaction if 83.6 liters of acetylene entered the reaction, and the practical yield was 80% of the theoretically possible?

Answer: 131, З6г

17. How much benzene is required to obtain 738 g of nitrobenzene, if the practical yield is 92% of the theoretical?

Answer 508.75g

1 8.Nitriding 46.8 of benzene yielded 66.42 g of nitrobenzene. Determine the practical yield of nitrobenzene in% of the theoretically possible.

19. How many grams of benzene can be obtained from 22.4 liters of acetylene, if the practical benzene yield is 40%.?

20. What volume of benzene (ρ \u003d 0.9 g / cm 3) is required to obtain 30.75 g of nitrobenzene if the nitration yield is 90% of the theoretically possible?

21. From 32 g of ethylene, 44 g of alcohol was obtained. Calculate the practical product yield in% of the theoretically possible.

22. How many grams of ethyl alcohol can be obtained from 1m 3 of natural gas containing 6% ethylene, if the practical yield is 80%?

23. What amount of acid and alcohol is needed to obtain 29.6 g of ethyl acetate if its yield is 80% of the theoretically possible?

24. When hydrolysis of 500 kg of wood, containing 50% cellulose, 70 kg of glucose is obtained. Calculate its practical yield in% of the theoretically possible.

25. How much glucose is obtained from 250 kg of sawdust containing 40% glucose. What amount of alcohol can be obtained from this amount of glucose at 85% practical yield?

Answer: 43.43g

26. How many grams of nitrobenzene should be taken to obtain 186 g of aniline by reduction, the yield of which is 92% of theoretical 27. Calculate the mass of the ester, which was obtained from 460 g formic acid and 460 g of ethyl alcohol. Ether yield from the theoretically possible is 80%.

28. When processing 1 ton of phosphorite containing 62% calcium phosphate, sulfuric acid was obtained 910.8 kg of superphosphate. Determine the output of superphosphate in% in relation to the theoretical.

Ca 3 (PO 4) 2 + 2H 2 S 0 4 \u003d Ca (H 2 P0 4) 2 + 2CaS 0 4

30. To obtain calcium nitrate, 1 ton of chalk was treated with dilute nitric acid. The yield of calcium nitrate was 85% in relation to the theoretical one. How much saltpeter was received?

Answer: 1394kg

31. 48 kg of ammonia were synthesized from 56 kg of nitrogen. What is the yield of ammonia as a percentage of theoretical.

Answer: 70.5%

32. 34 kg of ammonia was passed through a sulfuric acid solution. The yield of ammonium sulfate was 90% of theoretical. How many kilograms of ammonium sulfate have been obtained?

Answer: 118.8kg

Z3. In the oxidation of Z4 kg of ammonia, 54 kg of nitrogen oxide (II) were obtained. Calculate the yield of nitrogen oxide in% in relation to the theoretical.

34. In the laboratory, ammonia is produced by the interaction of ammonium chloride with slaked lime. How many grams of ammonia were obtained if 107 g of ammonium chloride were consumed and the ammonia yield was 90% of the theoretical?

Answer: 30.6g

35. From 60 kg of hydrogen and a corresponding amount of nitrogen, 272 kg of ammonia were synthesized. What is the yield of ammonia in% of the theoretically possible?

36. From 86.7 g of sodium nitrate containing 2% impurities, 56.7 g of nitric acid were obtained, what is the yield of nitric acid in% of the theoretically possible?

Answer: 90%.

37. By passing ammonia through 63kg of 50% nitric acid solution, 38kg of ammonium nitrate was obtained. What is its output in% to the theoretically possible?

38. For the production of phosphoric acid, 3I4kg of phosphorite containing 50% calcium phosphate was consumed. The phosphoric acid yield was 95%. How much acid was obtained?

Answer: 94, Zkg

39. 49 kg of a 50% sulfuric acid solution was neutralized with slaked lime, and 30.6 kg of calcium sulfate were obtained. Determine the product yield in% to the theoretical.

40. Phosphorus is obtained in technology according to the reaction equation;

Саз (Р0 4) 2 + 3SiО 2 + 5С → ЗСaSi О 3 + 2Р + 5СО

What is the yield of phosphorus in% of the theoretical, if it was obtained 12.4 kg from 77 kg of calcium phosphate?

Answer: 80.5%

41. Calculate the yield of calcium carbide in% to the theoretical, if 15.2 kg of it

were obtained from I4kg of calcium oxide.

42. Acetylene is obtained by interaction of calcium carbide with water

CaC 2 + 2H 2 0 \u003d Ca (OH) 2 + C 2 H 2

How many grams of acetylene will be obtained if 33.7 g of calcium carbide containing 5% of impurities are consumed and the acetylene yield is 90% of the theoretical?

Answer: 11.7g

43 in action of hydrochloric acid for 50 g of chalk, 20 g of carbon dioxide turned out. What is its output in% to the theoretical one?

Answer: 90.9%

44. When firing 1 ton of limestone containing 10% of impurities, the yield of carbon dioxide was 95%. How many kilograms of carbon dioxide were produced?

Answer: 376.2 kg.

45. Determine the yield of sodium silicate in% to the theoretical, if by fusing 10 kg of sodium hydroxide with sand, 12.2 kg of sodium silicate is obtained.

Element).

Express the mass value from the formula for the mass fraction of the substance: w \u003d m (x) * 100% / m, where w is the mass fraction of the substance, m (x) is the mass of the substance, m is the mass of the solution in which the substance is dissolved. To find the mass of a substance you need: m (x) \u003d w * m / 100%.

Calculate the mass you need from the formula for the product yield: product yield \u003d mp (x) * 100% / m (x), where mp (x) is the mass of the product x obtained in the real process, m (x) is the calculated mass of the substance x. Output: mp (x) \u003d product yield * m (x) / 100% or m (x) \u003d mp (x) * 100% / product yield. Given the output of the product in the condition of the problem, this formula will be necessary. If the yield is not given, then it should be considered 100%.

If the condition contains a reaction equation, then solve the problem using it. To do this, first make up the reaction equation, then calculate from it the amount of substance obtained or consumed for this reaction and substitute this amount of substance into the required formulas. For example, Na2SO4 + BaCl2 \u003d BaSO4 + 2NaCl. It is known that the mass of BaCl2 is 10.4 g, you need to find the mass of NaCl. Calculate the amount of barium chloride substance: n \u003d m / M. M (BaCl2) \u003d 208 g / mol. n (BaCl2) \u003d 10.4 / 208 \u003d 0.05 mol. It follows from the reaction equation that from 1 mol of BaCl2 2 mol of NaCl was formed. Calculate the amount of substance formed from 0.05 mol of BaCl2. n (NaCl) \u003d 0.05 * 2/1 \u003d 0.1 mol. In the problem, it was required to find the mass of sodium chloride, find it, having previously calculated the molar mass of sodium chloride. M (NaCl) \u003d 23 + 35.5 \u003d 58.5 g / mol. m (NaCl) \u003d 0.1 * 58.5 \u003d 5.85 g. The problem is solved.

note

Mass units can be milligrams, grams, kilograms.

Sources:

• "A manual on chemistry", G.P. Khomchenko, 2005.

The mass of a body is one of its most important physical characteristics, which shows its gravitational properties. Knowing the volume of a substance, as well as its density, one can easily calculate the mass of the body, which is based on this substance.

You will need

• The volume of matter V, its density p.

Instructions

Let us be given a heterogeneous one with mass V and mass m. Then it can be calculated using the formula:
p \u003d m / V.
From this it follows that in order to calculate the mass, you can use its consequence:
m \u003d p * V. Consider: Let us be given a platinum bar. Its volume is 6 cubic meters. Let's find its mass.
The task is solved in 2 steps:
1) According to the density table of various substances, the density of platinum is 21,500 kg / cu. meters.
2) Then, knowing the density and volume of this substance, we calculate its mass:
6 * 21500 \u003d 129000 kg, or 129 tons.

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Water, like any liquid, cannot always be weighed on a scale. But learn the mass water it is sometimes necessary both in some industries and in ordinary everyday situations, from calculating reservoirs to deciding how much water you can take with you in a kayak or rubber boat. In order to calculate the mass of water or any liquid placed in a particular volume, you first need to know its density.

You will need

• Volumetric dishes
• Ruler, tape measure or any other measuring device
• Water transfer vessel

Instructions

If you need to calculate the mass of water in a small vessel, you can do this using a conventional balance. Weigh the vessel together with first. Then transfer the water to another bowl. Then weigh the empty vessel. Subtract the empty mass from the full vessel. This will be the water contained in the vessel. Thus, it is possible to mass not only liquid, but also bulk substances, if it is possible to pour them into other dishes. This method can sometimes still be observed in some stores that do not have modern equipment. The seller first weighs an empty can or bottle, then fills it with sour cream, weighs it again, determines the weight of the sour cream, and only then calculates its value.

In order to determine the mass of water in a vessel that cannot be weighed, it is necessary to know two parameters - the density of water (or any other liquid) and the volume of the vessel. The density of the water is 1 g / ml. The density of another liquid can be found in a special table, which is usually found in reference books on chemistry.

If there is no measuring container to pour water into, calculate the volume of the container it is in. The volume is always equal to the product of the base area by the height, and usually there are no problems with vessels of a stable shape. The volume of water in the jar will be equal to the area of \u200b\u200bthe round base to the height filled with water. By multiplying the density? for the volume of water V, you get the mass of water m: m \u003d? * V.

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note

You can determine the mass also knowing the amount of water and its molar mass. The molar mass of water is 18, since it consists of the molar masses of 2 hydrogen atoms and 1 oxygen atom. MH2O \u003d 2MH + MO \u003d 21 + 16 \u003d 18 (g / mol). m \u003d n * M, where m is the mass of water, n is the amount, M is the molar mass.

Density is the ratio of mass to the volume it occupies - for solids, and the ratio of molar mass to molar volume - for gases. In its most general form, volume (or molar volume) will be the ratio of mass (or molar mass) to its density. The density is known. What to do? First determine the mass, then calculate the volume, then make the necessary corrections.

Instructions

Gas volume equal to ratio product multiplied by his - to the already known density. Others, even knowing, need to know the molar mass of the gas and the amount, that is, how many mole of gas you have. In principle, knowing how many moles of gas you have, you can calculate its volume without even knowing the density - according to Avogadro's law, one mole of any gas takes a volume of 22.4 liters. If it is imperative to calculate the volume through density, then you need to find out the mass of the gas in a volume that is still unknown.

Volume solid can be determined without even knowing the density, simply by measuring it, and in the case of a complex and very irregular shape the volume is determined, for example, by the volume of the liquid displaced by a solid. However, if it is necessary to calculate the volume precisely through the density, then the volume of a solid is the ratio of the body's mass to its density, and the mass is usually determined by simple weighing. If it is impossible to weigh the body for some reason (for example, it is too large or is moving), then you will have to resort to rather complicated indirect calculations. For example, for a moving body, mass is the ratio of twice the kinetic energy to the square of its velocity, or the ratio of the force applied to the body to its acceleration. For a very large body at rest, you will have to resort to calculations in relation to the mass of the Earth, using the gravitational constant and the moment of rotation. Or - through calculating the specific heat capacity of a substance; in any case, knowing only the density to calculate the volume will not be enough.

By calculating the mass of a solid, you can calculate volume by simply dividing mass by density.

note

1. The above methods are more or less applicable only in the case of homogeneity of the substance of which the solid consists
2. The above methods are more or less applicable in a relatively narrow temperature range - from minus 25 to plus 25 degrees Celsius. When it changes aggregate state substance density can change abruptly; in this case, formulas and calculation methods will be completely different.

Mass as a physical quantity is a parameter that characterizes the force of the body's action on gravity. To calculate body weight in physics, you need to know two of its values: the density of the body material and its volume.

Instructions

Let some body of volume V its p be given. Then it is calculated as follows:
m \u003d p * V. For clarity, the following is given:
Let an aluminum volume of 5 cubic meters be given. meters. The density of aluminum is 2700 kg / cu. meter. In this case, the mass of the bar will be:
m \u003d 2700/5 \u003d 540 kg.

note

The concept of mass is often confused with another, no less rare, physical quantity - weight. Weight is measured in N / m³ and characterizes the force that acts on the fulcrum. Mass, by its nature, does not have any point of support, and affects, as noted, only the gravity of the Earth.

Tip 6: How to find mass when volume and density are known

The mass of a body is its most important physical characteristic. In modern physical science there is a distinction between the concept of "mass": gravitational mass (as the degree of influence of a body on the earth's gravity) and inertial mass (what effort is required to bring the body out of the state of inertia). In any case, it is very easy to find the mass if the density and volume of the body are known.

Instructions

For clarity, you can give. It is required to find the mass of a concrete slab, whose volume is 15 m³.
Solution: the mass of a concrete slab is only required to know its density. In order to find out this information, you need to use the table of densities of various substances.

According to this table the concrete density is 2300 kg / m³. Then, in order to find the mass of a concrete slab, you will need to perform a simple algebraic action: m \u003d 15 * 2300 \u003d 34500 kg, or 34.5 tons. Answer: The mass of the concrete slab is 34.5 tons

Measurement of mass in the traditional way is carried out using one of the oldest instruments of mankind - with the help of scales. This is due to the comparison of body weight using a reference weight of the load - weights.

note

When calculating according to the above formula, it is necessary to realize that in this way the rest mass of a given body is recognized. An interesting fact is that many elementary particles have an oscillating mass, which depends on the speed of their movement. If elementary particle moves with the speed of the body, then this particle is massless (for example, a photon). If the speed of the particle is lower than the speed of light, then such a particle is called massive.

Useful advice

When measuring mass, one should never forget in which system the final result will be given. It means that in the SI system, mass is measured in kilograms, while in the CGS system, mass is measured in grams. Also, mass is measured in tons, centners, carats, pounds, ounces, poods, as well as in many other units, depending on the country and culture. In our country, for example, since ancient times, the mass was measured in poods, berkovtsy, zolotniks.

Sources:

• concrete slab weight

All substances have a certain density. The density is calculated depending on the volume occupied and the target mass. It is found on the basis of experimental data and numerical transformations. In addition, the density depends on many different factors, due to which its constant value changes.

Instructions

Imagine that you are given a vessel filled to the brim with water. In the problem, it is necessary to find the density of water, while not knowing either mass or volume. In order to calculate the density, both parameters must be found experimentally. Start by determining the mass.
Take the vessel and place it on the scale. Then pour the water out of it, and then put the vessel back on the same scales. Compare the measurement results and get the formula for finding the mass of water:
mb.- mw. \u003d mv., where mob. is the mass of the vessel with water (total mass), mс is the mass of the vessel without water.
The second thing you need to find is water. Pour water into a measuring vessel, then use the scale on it to determine the volume of water contained in the vessel. Only after that, using the formula, find the density of water:
ρ \u003d m / V
This experiment can only roughly determine the density of water. However, under the influence of some factors, it can. Check out the most important of these factors.

At a water temperature of t \u003d 4 ° C, water has a density ρ \u003d 1000 kg / m ^ 3 or 1 g / cm ^ 3. With a change, the density also changes. In addition, pressure, salinity and salinity are among the factors affecting density. The most pronounced effect on the density of temperature.
Remember that density under the influence of temperature changes in a parabolic manner. The value t \u003d 4 ° C is the critical point of this parabola, at which the water density reaches its highest value. Any temperature above or below this value leads to a decrease in density. At 0 ° C, the density of the water decreases significantly.

Mineralization and pressure affect the density of water in the same way. When they increase, the density increases. Also, the noticeable density of water is directly proportional to the concentration of salt in it.
There are other factors on which the density of water depends, but their influence is much weaker than that of the above.

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The mass of a substance is the measure by which the body acts on its support. It is measured in kilograms (kg), grams (g), tons (t). It is very easy to find the mass of a substance if its volume is known.

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When teaching students to solve computational problems in chemistry, teachers face a number of problems

• solving a problem, students do not understand the essence of the problem and the course of their solution;
• do not analyze the content of the problem;
• do not determine the sequence of actions;
• misuse of chemical language, mathematical operations and notation physical quantities and etc.;

Overcoming these shortcomings is one of the main goals that a teacher sets for himself when starting to learn how to solve computational problems.

The teacher's task is to teach students to analyze the conditions of problems, through drawing up a logical scheme for solving a specific problem. Logging the problem prevents many of the mistakes students make.

Lesson objectives:

• formation of the ability to analyze the condition of the problem;
• formation of the ability to determine the type of computational problem, the procedure for solving it;
• development of cognitive, intellectual and creative abilities.

Lesson objectives:

• master ways of solving chemical tasks using the concept “mass fraction of the reaction product yield from theoretical”;
• to develop skills in solving computational problems;
• facilitate the assimilation of material related to production processes;
• stimulate in-depth study of theoretical issues, interest in solving creative problems.

During the classes

We determine the reason and essence of the situation, which are described in the tasks “to get the product out of the theoretical”.

In real chemical reactions, the mass of the product is always less than the calculated one. Why?

• Many chemical reactions are reversible and never go to completion.
• When interacting organic matter by-products are often formed.
• In heterogeneous reactions, substances mix poorly, and some of the substances simply do not react.
• Some of the gaseous substances may escape.
• When precipitation is received, part of the substance may remain in solution.

Conclusion:

• theoretical mass is always more practical;
• the theoretical volume is always greater than the practical volume.

The theoretical yield is 100%, the practical yield is always less than 100%.

The amount of product calculated according to the reaction equation - theoretical yield, corresponds to 100%.

The fraction of the yield of the reaction product (- "etta") is the ratio of the mass of the resulting substance to the mass that should have been obtained in accordance with the calculation according to the reaction equation.

Three types of tasks with the concept of “product output”:

1. Given the masses starting materialand reaction product... Determine the product yield.

2. Given the masses starting materialand exit reaction product.Determine the mass of the product.

3. Given the masses product and exit product. Determine the mass of the starting material.

Tasks.

1. When burning iron in a vessel containing 21.3 g of chlorine, 24.3 g of iron (III) chloride were obtained. Calculate the yield of the reaction product.

2. Hydrogen was passed over 16 g of sulfur when heated. Determine the volume (NL) of the obtained hydrogen sulfide if the yield of the reaction product is 85% of the theoretically possible.

3. What volume of carbon monoxide (II) was taken to reduce iron oxide (III), if 11.2 g of iron was obtained with a yield of 80% of the theoretically possible.

Analysis of tasks.

Each task is made up of a set of data (known substances) - the conditions of the problem (“exit”, etc.) - and a question (substances whose parameters are to be found). In addition, it has a system of dependencies that connect the desired with the data and the data with each other.

Analysis tasks:

1) reveal all the data;

2) identify dependencies between data and conditions;

3) identify the relationship between the data and the desired.

So, we find out:

1. What substances are we talking about?

2. What changes have occurred with the substances?

3. What quantities are named in the problem statement?

4. What data - practical or theoretical, are named in the problem statement?

5. Which of the data can be directly used to calculate the reaction equations, and which must be converted using the mass fraction of the output?

Algorithms for solving problems of three types:

Determination of the product yield in% of the theoretically possible.

1. Write down the equation of the chemical reaction and place the coefficients.

2. Under the formulas of substances write the amount of substance according to the coefficients.

3. The mass obtained in practice is known.

4. Determine the theoretical mass.

5. Determine the yield of the reaction product (%), referring the practical mass to the theoretical and multiplying by 100%.

6. Write down your answer.

Calculation of the mass of the reaction product if the product yield is known.

1. Write down "given" and "find", write down the equation, place the coefficients.

2. Find the theoretical amount of substance for the starting substances. n \u003d

3. Find the theoretical amount of the substance of the reaction product, according to the coefficients.

4. Calculate the theoretical mass or volume of the reaction product.

m \u003d M * n or V \u003d V m * n

5. Calculate the practical mass or volume of the reaction product (multiply the theoretical mass or theoretical volume by the yield fraction).

Calculation of the mass of the starting material if the mass of the reaction product and the product yield are known.

1. From a known practical volume or mass, find the theoretical volume or mass (using the yield fraction).

2. Find the theoretical amount of substance for the product.

3. Find the theoretical amount of substance for the original substance, according to the coefficients.

4. Using the theoretical amount of a substance, find the mass or volume of the starting materials in the reaction.

Homework.

Solve the tasks:

1. For the oxidation of sulfur oxide (IV) took 112 l (NU) oxygen and received 760 g of sulfur oxide (VI). What is the yield of the product as a percentage of the theoretically possible?

2. The interaction of nitrogen and hydrogen received 95 g of ammonia NH 3 with a yield of 35%. What volumes of nitrogen and hydrogen were taken for the reaction?

3. 64.8 g of zinc oxide was reduced with excess carbon. Determine the mass of the formed metal if the yield of the reaction product is 65%.