Question

СРОЧНО ХИМИЯ ПЖ! 1)Визначте і вкажіть невідомий вуглеводень у реакції: 2Х + 11О2 → 8СО2 + 6Н2О 2) Обчисліть вихід продукту реакції від теоретично можливого (у відсотках), якщо з хлор метану масою 40,4 г за допомогою реакції Вюрца було добуто етан об’ємом 6,72 л. 3) Визначте і вкажіть масу нітратної кислоти, необхідну для одержання нітробензену, маса якого становить 49,2 г. 4) Виконайте перетворення: СН4 → С2Н2 → С6Н6 → С6H5Cl → C6H4ClNO2

Answer 1

Ответ:

1)2C4H6+1102-->8C02+6H20

2) фото 1

3)фото 2

4) фото 3

Answer 2

Unknown hydrocarbon in the reaction: 2X + 11O2 -> 8CO2 + 6H2O

To determine the unknown hydrocarbon in the given reaction, we need to balance the equation first. Balancing the equation will help us determine the stoichiometric coefficients of the reactants and products.

The balanced equation for the reaction is:

2X + 11O2 -> 8CO2 + 6H2O

From the balanced equation, we can see that for every 2 moles of X, we obtain 8 moles of CO2 and 6 moles of H2O.

Unfortunately, without additional information, it is not possible to determine the specific identity of the hydrocarbon X. The reaction equation does not provide enough information about the structure or properties of the hydrocarbon. To determine the exact identity of X, additional information or experimental data is needed.

Calculation of the theoretical yield of the reaction

To calculate the theoretical yield of the reaction, we need to know the balanced equation and the amount of the limiting reactant used.

The balanced equation for the reaction is:

CH4 + Cl2 -> CH3Cl + HCl

From the equation, we can see that 1 mole of CH4 reacts with 1 mole of Cl2 to produce 1 mole of CH3Cl.

Given: - Mass of CH4 = 40.4 g - Volume of CH3Cl = 6.72 L

To calculate the theoretical yield, we need to convert the mass of CH4 to moles using its molar mass. The molar mass of CH4 is approximately 16.04 g/mol.

Number of moles of CH4 = Mass of CH4 / Molar mass of CH4 Number of moles of CH4 = 40.4 g / 16.04 g/mol ≈ 2.52 mol

Since the balanced equation shows a 1:1 mole ratio between CH4 and CH3Cl, the number of moles of CH3Cl produced will also be 2.52 mol.

To calculate the volume of CH3Cl, we need to use the ideal gas law equation:

PV = nRT

Assuming the reaction takes place at standard temperature and pressure (STP), we can use the values: - Pressure (P) = 1 atm - Temperature (T) = 273 K - Gas constant (R) = 0.0821 L·atm/(mol·K)

Volume of CH3Cl = nRT / P Volume of CH3Cl = 2.52 mol * 0.0821 L·atm/(mol·K) * 273 K / 1 atm ≈ 57.7 L

The theoretical yield of the reaction is approximately 57.7 L of CH3Cl.

Calculation of the mass of nitric acid required to obtain nitrobenzene

To calculate the mass of nitric acid required to obtain nitrobenzene, we need to know the balanced equation and the molar mass of nitric acid (HNO3) and nitrobenzene (C6H5NO2).

The balanced equation for the reaction is:

C6H6 + HNO3 -> C6H5NO2 + H2O

From the equation, we can see that 1 mole of C6H6 reacts with 1 mole of HNO3 to produce 1 mole of C6H5NO2.

Given: - Mass of nitrobenzene (C6H5NO2) = 49.2 g

To calculate the number of moles of nitrobenzene, we need to divide its mass by its molar mass. The molar mass of nitrobenzene is approximately 123.11 g/mol.

Number of moles of C6H5NO2 = Mass of C6H5NO2 / Molar mass of C6H5NO2 Number of moles of C6H5NO2 = 49.2 g / 123.11 g/mol ≈ 0.399 mol

Since the balanced equation shows a 1:1 mole ratio between C6H6 and C6H5NO2, the number of moles of HNO3 required will also be 0.399 mol.

To calculate the mass of HNO3, we need to multiply the number of moles of HNO3 by its molar mass. The molar mass of HNO3 is approximately 63.01 g/mol.

Mass of HNO3 = Number of moles of HNO3 * Molar mass of HNO3 Mass of HNO3 = 0.399 mol * 63.01 g/mol ≈ 25.14 g

The mass of nitric acid required to obtain nitrobenzene is approximately 25.14 g.

Transformation: CH4 -> C2H2 -> C6H6 -> C6H5Cl -> C6H4ClNO2

The given transformation involves a series of reactions starting from methane (CH4) and ending with 2-chloro-4-nitroaniline (C6H4ClNO2).

1. CH4 -> C2H2: - This reaction is known as the pyrolysis or cracking of methane. - It involves the breaking of carbon-hydrogen bonds in methane to form carbon-carbon double bonds in ethyne (acetylene). - The reaction can be represented as: CH4 -> C2H2 + H2 - The conditions required for this reaction include high temperature and the absence of oxygen. - The reaction is endothermic. - The catalysts such as alumina or silica are often used to increase the reaction rate. - The balanced equation for this reaction is: CH4 -> C2H2 + 2H2

2. C2H2 -> C6H6: - This reaction is known as the polymerization of acetylene. - It involves the combination of multiple acetylene molecules to form benzene. - The reaction can be represented as: 3C2H2 -> C6H6 - The reaction is exothermic. - The reaction is typically carried out under high pressure and in the presence of a catalyst, such as iron or nickel. - The balanced equation for this reaction is: 3C2H2 -> C6H6

3. C6H6 -> C6H5Cl: - This reaction is known as the chlorination of benzene. - It involves the substitution of one hydrogen atom in benzene with a chlorine atom. - The reaction can be represented as: C6H6 + Cl2 -> C6H5Cl + HCl - The reaction is typically carried out in the presence of a catalyst, such as an iron halide or aluminum chloride. - The balanced equation for this reaction is: C6H6 + Cl2 -> C6H5Cl + HCl

4. C6H5Cl -> C6H4ClNO2: - This reaction is known as the nitration of chlorobenzene. - It involves the substitution of one hydrogen atom in chlorobenzene with a nitro group (-NO2). - The reaction can be represented as: C6H5Cl + HNO3 -> C6H4ClNO2 + H2O - The reaction is typically carried out in the presence of a catalyst, such as concentrated sulfuric acid. - The balanced equation for this reaction is: C6H5Cl + HNO3 -> C6H4ClNO2 + H2O