При нагревании до некоторой температуры 36 г уксусной кислоты и 7.36 г 100%-ного этанола в присутствии серной кислоты получена равновесная смесь. Эта смесь при действии избытка раствора хлорида бария образует 4.66 г осадка, а при действии избытка раствора гидрокарбоната калия выделяет 12.1 л диоксида углерода (н.у.). Найдите количество сложного эфира (в молях) в равновесной смеси, которая образуется при нагревании до такой же температуры 150 г уксусной кислоты и 200 мл 90%-ного этанола (плотность 0.82 г/мл) в присутствии серной кислоты в качестве катализатора.Как решить эту задачу?
ответ: 0.8 моль
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Если не знаешь что в продуктах то попробуй написать в гугле Fischer esterification
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@eudaimon помоги пожалуйста ему решить задачу🥹
To solve this problem, we must first understand the chemical process involved, which is the Fischer esterification reaction. This reaction occurs between an alcohol (ethanol) and a carboxylic acid (acetic acid) to form an ester (ethyl acetate) and water.
The chemical equations for the reactions with barium chloride and potassium bicarbonate are as follows:
- Reaction with barium chloride forms barium sulfate as a precipitate:
\text{CH}_3\text{COOH} + \text{BaCl}_2 \rightarrow \text{BaSO}_4 \downarrow + 2 \text{CH}_3\text{COCl}
- Reaction with potassium bicarbonate releases carbon dioxide gas:
2 \text{CH}_3\text{COOH} + 2 \text{KHCO}_3 \rightarrow 2 \text{CO}_2 \uparrow + 2 \text{H}_2\text{O} + 2 \text{KCH}_3\text{COO}
We will use stoichiometry to relate the given masses and volumes to the amount of ester formed.
First, let’s find the moles of barium sulfate and carbon dioxide to determine the moles of acetic acid that reacted:
For barium sulfate:
m(\text{BaSO}_4) = 4.66 \, \text{g}
Using the molar mass of barium sulfate, M(\text{BaSO}_4) = 233.39 \, \text{g/mol} , we can find the moles:
n(\text{BaSO}_4) = \frac{m}{M} = \frac{4.66 \, \text{g}}{233.39 \, \text{g/mol}}
For carbon dioxide:
Standard conditions imply that 1 mole of gas occupies 22.4 liters. Therefore:
V(\text{CO}_2) = 12.1 \, \text{L}
n(\text{CO}_2) = \frac{V}{22.4 \, \text{L/mol}} = \frac{12.1 \, \text{L}}{22.4 \, \text{L/mol}}
Next, we’ll find the moles of acetic acid and ethanol in the second part of the problem:
For acetic acid:
m(\text{CH}_3\text{COOH}) = 150 \, \text{g}
Knowing the molar mass of acetic acid, M(\text{CH}_3\text{COOH}) = 60.05 \, \text{g/mol} , we can find the moles:
n(\text{CH}_3\text{COOH}) = \frac{m}{M} = \frac{150 \, \text{g}}{60.05 \, \text{g/mol}}
For ethanol:
Given the density \rho = 0.82 \, \text{g/mL} and volume V = 200 \, \text{mL} , the mass of ethanol is:
m(\text{C}_2\text{H}_5\text{OH}) = \rho \cdot V = 0.82 \, \text{g/mL} \cdot 200 \, \text{mL}
But since the solution is 90% ethanol by mass, the mass of pure ethanol is:
m(\text{C}_2\text{H}_5\text{OH, pure}) = 0.90 \cdot m(\text{C}_2\text{H}_5\text{OH})
Using the molar mass of ethanol, M(\text{C}_2\text{H}_5\text{OH}) = 46.07 \, \text{g/mol} , we can find the moles of pure ethanol:
n(\text{C}_2\text{H}_5\text{OH, pure}) = \frac{m(\text{C}_2\text{H}_5\text{OH, pure})}{M(\text{C}_2\text{H}_5\text{OH})}
Now calculate the moles of each substance using the equations above, and then relate them to the initial conditions to find the mole ratio and thus determine the moles of ethyl acetate formed in the equilibrium mixture. The ratio of moles of acetic acid that reacted to form ester in the original mixture will be the same as in the larger mixture if we assume the reaction conditions and the extent of the reaction are identical.
Apply the mole ratio to the moles of acetic acid and ethanol in the larger mixture to find the moles of ester formed. Remember that the stoichiometry of esterification is 1:1 for acid to alcohol, so the limiting reactant will determine the maximum amount of ester that can be formed.
I have provided you with the framework to solve the problem. You must now perform the calculations to find the answer.
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