Question

Помогите , пожалуйста, очень срочно!!! Осуществите СН3-CH2OH->Ch3-CH=O->CH3-COOH->(CH3-COO)2Ca->CH3-COONa Какой объем метана(н.у.) можно полностью окислить 448м3 воздуха?

Answer 1

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Answer 2

Conversion of Methanol to Acetic Acid

The given reaction sequence involves the conversion of methanol (CH3OH) to acetic acid (CH3COOH) through several steps. Let's break down the reaction sequence step by step:

1. SN2 Reaction: The first step involves the conversion of methanol (CH3OH) to formaldehyde (CH2O) through a substitution reaction with a strong base like sodium hydroxide (NaOH). This reaction is known as an SN2 reaction. The reaction can be represented as follows: CH3OH + NaOH → CH2O + Na+ + H2O

2. Oxidation of Formaldehyde: The next step is the oxidation of formaldehyde (CH2O) to formic acid (HCOOH) using an oxidizing agent like potassium dichromate (K2Cr2O7) and sulfuric acid (H2SO4). The reaction can be represented as follows: CH2O + H2SO4 + K2Cr2O7 → HCOOH + Cr2(SO4)3 + K2SO4 + H2O

3. Oxidation of Formic Acid: The third step involves the oxidation of formic acid (HCOOH) to acetic acid (CH3COOH) using the same oxidizing agent, potassium dichromate (K2Cr2O7), and sulfuric acid (H2SO4). The reaction can be represented as follows: HCOOH + H2SO4 + K2Cr2O7 → CH3COOH + Cr2(SO4)3 + K2SO4 + H2O

4. Formation of Calcium Acetate: The fourth step involves the reaction of acetic acid (CH3COOH) with calcium hydroxide (Ca(OH)2) to form calcium acetate [(CH3COO)2Ca] and water (H2O). The reaction can be represented as follows: 2CH3COOH + Ca(OH)2 → (CH3COO)2Ca + 2H2O

5. Formation of Sodium Acetate: The final step involves the reaction of calcium acetate [(CH3COO)2Ca] with sodium carbonate (Na2CO3) to form sodium acetate (CH3COONa), calcium carbonate (CaCO3), and water (H2O). The reaction can be represented as follows: (CH3COO)2Ca + Na2CO3 → 2CH3COONa + CaCO3

Calculation of Methane Volume

To calculate the volume of methane (CH4) required to completely oxidize 448 m3 of air at standard conditions (n.у.), we need to consider the stoichiometry of the reaction between methane and oxygen.

The balanced equation for the combustion of methane is as follows: CH4 + 2O2 → CO2 + 2H2O

From the balanced equation, we can see that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.

To calculate the volume of methane, we need to convert the given volume of air (448 m3) to the number of moles of oxygen. Since air is approximately 21% oxygen, we can calculate the moles of oxygen using the ideal gas law.

The ideal gas law equation is given by: PV = nRT

Where: P = pressure (atm) V = volume (m3) n = number of moles R = ideal gas constant (0.0821 L·atm/(mol·K)) T = temperature (K)

At standard conditions (n.у.), the pressure is 1 atm and the temperature is 273 K.

Using the ideal gas law, we can calculate the number of moles of oxygen: n = PV / RT

Assuming the pressure and temperature are constant, we can simplify the equation to: n = V / 22.4

Since 1 mole of methane reacts with 2 moles of oxygen, the number of moles of methane required will be half of the number of moles of oxygen.

Finally, we can calculate the volume of methane using the number of moles and the molar volume of methane at standard conditions (n.у.), which is 22.4 L/mol.

Let's perform the calculations:

1. Calculate the number of moles of oxygen: n_oxygen = V_air / 22.4

2. Calculate the number of moles of methane: n_methane = n_oxygen / 2

3. Calculate the volume of methane: V_methane = n_methane * 22.4

Substituting the given volume of air (448 m3) into the equation, we can calculate the volume of methane.

Please note that the above calculations assume ideal gas behavior and do not take into account any other factors that may affect the reaction.

I hope this helps! Let me know if you have any further questions.