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

Рассчитаем какое из двух данных веществ взято в избытке, а потом решаем по недостатку
 nC2H5OH=11.5/46=0.25mol в избытке
 nCH3COOH=30/60=0.2mol
                    30g                  xg
С2H5OH+CH3COOH=CH3COOC2H5+H2O
                    60g                   88g

x=30·88/60=44g

mprakt=44·80%/100%=35.2%

Answer 2

Calculating the Amount of Ethyl Acetate

To calculate the amount of ethyl acetate that can be obtained from 11.5 g of ethanol and 30 g of acetic acid with an 80% yield, we can use the following steps:

1. Determine the Molar Mass of Ethanol and Acetic Acid: - The molar mass of ethanol (C2H5OH) is 46.07 g/mol. - The molar mass of acetic acid (CH3COOH) is 60.05 g/mol.

2. Convert the Given Masses to Moles: - Moles of ethanol = 11.5 g / 46.07 g/mol - Moles of acetic acid = 30 g / 60.05 g/mol

3. Identify the Limiting Reagent: - Use the mole ratio of the balanced chemical equation to determine which reactant limits the amount of product that can be formed.

4. Calculate the Theoretical Yield of Ethyl Acetate: - Once the limiting reagent is identified, use the stoichiometry of the balanced chemical equation to calculate the theoretical yield of ethyl acetate.

5. Apply the Yield Percentage: - Finally, apply the given yield percentage to find the actual amount of ethyl acetate produced.

Let's proceed with the calculations.

Step 1: Determine the Molar Mass

- Molar mass of ethanol (C2H5OH): 46.07 g/mol - Molar mass of acetic acid (CH3COOH): 60.05 g/mol

Step 2: Convert the Given Masses to Moles

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

Step 3: Identify the Limiting Reagent

- The balanced chemical equation for the reaction is: `CH3COOH + C2H5OH → CH3COOC2H5 + H2O` - From the equation, the mole ratio of acetic acid to ethanol is 1:1. - Since the moles of acetic acid (0.50 moles) are greater than the moles of ethanol (0.25 moles), ethanol is the limiting reagent.

Step 4: Calculate the Theoretical Yield of Ethyl Acetate

- The stoichiometry of the balanced chemical equation shows that 1 mole of ethanol produces 1 mole of ethyl acetate. - The theoretical yield of ethyl acetate from 0.25 moles of ethanol is also 0.25 moles.

Step 5: Apply the Yield Percentage

- Given yield percentage = 80% - Actual yield = Theoretical yield * Yield percentage - Actual yield = 0.25 moles * 80% = 0.20 moles

Conclusion

The amount of ethyl acetate that can be obtained from 11.5 g of ethanol and 30 g of acetic acid, with an 80% yield, is 0.20 moles.