Question

Сколько граммов этилового эфира уксусной кислоты можно получить из 11,5 г этанола и 30 г уксусной кислоты, если выход реакции составляет 80% от теоретического?Решение распишите подробно..

Answer 1

Для сокращения записи обозначу C₂H₅- как Et, а CH₃- как Me, т.е. этанол C₂H₅OH = EtOH, уксусная кислота CH₃COOH = MeCOOH, их эфир CH₃COOC₂H₅ = MeCOOEt.

Дано:
m(EtOH) = 11,5г
m(MeCOOH) = 30г
φ = 80% = 0,80
Найти:
m(MeCOOEt)пр.
Решение:
EtOH + MeCOOH = MeCOOEt + H₂O
Mr(C₂H₅OH) = 46 г/моль 
Mr(CH₃COOH) = 60 г/моль
Для реакции с 30г кислоты потребуется: (30×46)÷60 = 23г этанола, а по условию дано 11,5г, значит он в недостатке, считаем по нему массу эфира: 
Mr(CH₃COOC₂H₅) = 88 г/моль
m(MeCOOEt) = (11.5×88)÷46 = 22г
m(MeCOOEt)пр. = 22×0,80 = 17,6г
Ответ: m(MeCOOEt)пр. = 17,6г

Answer 2

Calculation of the Amount of Ethyl Acetate Produced

To calculate the amount of ethyl acetate that can be produced from 11.5 g of ethanol and 30 g of acetic acid, we need to consider the reaction between these two compounds. The reaction between ethanol and acetic acid produces ethyl acetate and water. The balanced chemical equation for this reaction is as follows:

CH3CH2OH + CH3COOH → CH3COOCH2CH3 + H2O

From the balanced equation, we can see that one mole of ethanol reacts with one mole of acetic acid to produce one mole of ethyl acetate and one mole of water.

To calculate the moles of ethanol and acetic acid, we need to divide their masses by their respective molar masses. The molar mass of ethanol (C2H5OH) is approximately 46.07 g/mol, and the molar mass of acetic acid (CH3COOH) is approximately 60.05 g/mol.

Moles of ethanol = mass of ethanol / molar mass of ethanol Moles of acetic acid = mass of acetic acid / molar mass of acetic acid

Let's calculate the moles of ethanol and acetic acid:

Moles of ethanol = 11.5 g / 46.07 g/mol Moles of acetic acid = 30 g / 60.05 g/mol

Now, we can determine the limiting reactant, which is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed. The limiting reactant is the one that produces the smallest number of moles of product.

To find the limiting reactant, we compare the moles of ethanol and acetic acid. The reactant with the smaller number of moles is the limiting reactant.

Let's compare the moles of ethanol and acetic acid:

Moles of ethanol = 11.5 g / 46.07 g/mol ≈ 0.25 mol Moles of acetic acid = 30 g / 60.05 g/mol ≈ 0.50 mol

From the calculations, we can see that ethanol is the limiting reactant because it produces fewer moles of product compared to acetic acid.

Next, we need to determine the theoretical yield of ethyl acetate, which is the maximum amount of ethyl acetate that can be produced from the limiting reactant. The theoretical yield is calculated based on the stoichiometry of the balanced equation.

From the balanced equation, we can see that one mole of ethanol reacts to produce one mole of ethyl acetate. Therefore, the theoretical yield of ethyl acetate is equal to the moles of ethanol.

Theoretical yield of ethyl acetate = moles of ethanol ≈ 0.25 mol

However, the problem states that the yield of the reaction is only 80% of the theoretical yield. To calculate the actual yield of ethyl acetate, we multiply the theoretical yield by the yield percentage.

Actual yield of ethyl acetate = theoretical yield * yield percentage Actual yield of ethyl acetate = 0.25 mol * 0.80 ≈ 0.20 mol

Finally, we can calculate the mass of ethyl acetate produced by multiplying the actual yield by the molar mass of ethyl acetate.

Mass of ethyl acetate = actual yield of ethyl acetate * molar mass of ethyl acetate

The molar mass of ethyl acetate (CH3COOCH2CH3) is approximately 88.11 g/mol.

Mass of ethyl acetate = 0.20 mol * 88.11 g/mol ≈ 17.62 g

Therefore, approximately 17.62 grams of ethyl acetate can be produced from 11.5 grams of ethanol and 30 grams of acetic acid, assuming an 80% yield.

Answer:

Approximately 17.62 grams of ethyl acetate can be produced from 11.5 grams of ethanol and 30 grams of acetic acid, assuming an 80% yield.