Question

Какую массу этилацетата можно получить при взаимодейст¬вии этанола массой 1,61г и уксусной кислоты массой 1,80 г? Мас¬совая доля выхода эфира равна 75%.

Answer 1

1,61г------------1,8г----------------хг
C2H5OH + CH3COOH = CH3COOCH2CH3 + H2O
46г-------------60г-----------------88г
m (этанола) = 46 * 1,8 / 60 = 1,38 г надо этанола. он в избытке, считаем по кислоте.
m (эфира) = 1,8 * 88 / 60 = 2,64 г
доля выхода = m практич / m теоретич
0,75 = m / 2,64
m = 2,64 * 0,75
m = 1,98 г

Answer 2

Calculation of the Mass of Ethyl Acetate Produced

To calculate the mass of ethyl acetate produced, we need to consider the reaction between ethanol and acetic acid, which forms ethyl acetate and water. The balanced chemical equation for this reaction is:

C2H5OH + CH3COOH → CH3COOC2H5 + H2O

Given the mass of ethanol (1.61 g) and the mass of acetic acid (1.80 g), we can calculate the mass of ethyl acetate produced.

To do this, we need to determine the limiting reactant, which is the reactant that is completely consumed and determines the maximum amount of product that can be formed. The limiting reactant is the one that produces the least amount of product based on the stoichiometry of the reaction.

To find the limiting reactant, we need to compare the number of moles of each reactant. We can calculate the number of moles using the molar mass of each compound.

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.

Let's calculate the number of moles for each reactant:

- Moles of ethanol (C2H5OH) = mass of ethanol / molar mass of ethanol - Moles of acetic acid (CH3COOH) = mass of acetic acid / molar mass of acetic acid

Now, we can compare the moles of each reactant to determine the limiting reactant.

After determining the limiting reactant, we can use the stoichiometry of the reaction to calculate the theoretical yield of ethyl acetate. The stoichiometry tells us the ratio of moles between the reactants and products.

In this case, the stoichiometry of the reaction tells us that 1 mole of ethanol reacts with 1 mole of acetic acid to produce 1 mole of ethyl acetate.

Finally, we can calculate the mass of ethyl acetate produced using the theoretical yield and the given percentage yield.

Let's perform the calculations step by step.

Step 1: Calculate the number of moles for each reactant

- Moles of ethanol (C2H5OH) = 1.61 g / 46.07 g/mol = 0.0349 mol- Moles of acetic acid (CH3COOH) = 1.80 g / 60.05 g/mol = 0.0299 mol

Step 2: Determine the limiting reactant

To determine the limiting reactant, we compare the moles of each reactant. The reactant with the smaller number of moles is the limiting reactant.

In this case, acetic acid (CH3COOH) has fewer moles (0.0299 mol) compared to ethanol (C2H5OH) (0.0349 mol). Therefore, acetic acid is the limiting reactant.

Step 3: Calculate the theoretical yield of ethyl acetate

Since acetic acid is the limiting reactant, the number of moles of ethyl acetate produced is equal to the number of moles of acetic acid.

Therefore, the theoretical yield of ethyl acetate is 0.0299 mol.

Step 4: Calculate the mass of ethyl acetate produced

To calculate the mass of ethyl acetate produced, we multiply the number of moles by the molar mass of ethyl acetate.

The molar mass of ethyl acetate (CH3COOC2H5) is approximately 88.11 g/mol.

Mass of ethyl acetate = theoretical yield of ethyl acetate * molar mass of ethyl acetate Mass of ethyl acetate = 0.0299 mol * 88.11 g/mol = 2.636 g

Therefore, the mass of ethyl acetate that can be produced is approximately 2.636 grams.

Please note that the given percentage yield of 75% indicates that the actual yield of ethyl acetate obtained in the reaction is 75% of the theoretical yield. To calculate the actual yield, we would multiply the theoretical yield by the percentage yield.

However, the given information does not specify the actual yield, so we cannot calculate it with the given data.