Chemistry 9 Lesson 48: Practice Ethyl alcohol, acetic acid and fats

1. Theoretical Summary

a. Ethyl alcohol:

– Structural formula: ON₃– ONLY₂-OH

– Physical properties:

• It is a colorless, liquid that boils at 78.30, lighter than water
• Infinitely soluble in water, soluble in many substances such as iodine, benzene, etc.

– Chemical properties:

OLD₂H₆O + 3O₂ \(\overset{t^{0}}{\rightarrow}\) 2CO₂ + 3 HOURS₂O

2 C₂H₅OH + 2Na → 2C₂H₅ONa + FAMILY₂

• Reaction with acetic acid (esterification reaction)

OLD₂H₅OH + ONLY₃COOH ONLY₃COOC₂H₅ + FRIENDS₂O

b. Acetic acid

– Structural formula: \(\begin{array}{l} C{H_3} – \mathop C\limits_\parallel – OH\\ {\rm{ \,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,O}} \end{array}\)

– Physical properties: is a liquid, colorless, sour taste, infinitely soluble in water.

– Chemical properties:

– Acetic acid has the properties of a weak acid.

• Reaction with Ethyl Alcohol:

OLD₂H₅OH + ONLY₃COOH ONLY₃COOC₂H₅ + FRIENDS₂O

c. Fat

– Structural formula: (R-COO)₃OLD₃H₅

– Physical properties: lighter than water, insoluble in water, soluble in benzene, gasoline, kerosene, …

– Chemical properties:

• Hydrolysis in acidic medium.

(RCOO)₃OLD₃H₅ + 3 HOURS₂O \(\xrightarrow[axit]{t^{0}}\)C₃H₅(OH)₃ + 3RCOOH

• Hydrolysis in alkaline media (saponification reaction)

(RCOO)₃OLD₃H₅ + 3NaOH \(\overset{t^{0}}{\rightarrow}\) C₃H₅(OH)₃ + 3RCOONa

2. Illustrated exercise

2.1. Form 1: Chemical Equation

Fill in the question marks with the appropriate substances and write the chemical equations of the following reaction diagrams:

a) OLD₂H₅OH + ? → ? + FRIENDS₂\(\uparrow\)

b) OLD₂H₅OH + ? \(\overset{t^{0}}{\rightarrow}\) CO₂ + ?

c) ONLY₃COOH + ? ONLY₃COOC₂H₅ + ?

d) ONLY₃COOH + ? → (ONLY₃COO)₂Mg + ?

e) ONLY₃COOH + KOH → ? + ?

f) (RCOO)₃OLD₃H₅ + ? \(\overset{t^{0}}{\rightarrow}\) ? + RCOONa

g) ONLY₃COOC₂H₅ + FRIENDS₂O \(\xrightarrow[axit]{t^{0}}\) ? + ?

h)₂H₄(k) + ? \(\overset{Acid}{\rightarrow}\) OVER₂H₅OH

i) ? + Zn → (ONLY₃COO)₂Zn + ? \(\uparrow\)

k) 2 ONLY₃COOH + ? → (ONLY₃COO)₂Ca + ? + FRIENDS₂O

Solution guide

a) 2C₂H₅OH + 2Na → 2C₂H₅ONa + FAMILY₂\(\uparrow\)

b) OLD₂H₅OH + 3O₂ \(\overset{t^{0}}{\rightarrow}\) 2CO₂ + 3 HOURS₂O

c) ONLY₃COOH + C₂H₅OH ONLY₃COOC₂H₅ + FRIENDS₂O

d) 2 ONLY₃COOH + Mg → (ONLY₃COO)₂Mg + H₂\(\uparrow\)

e) ONLY₃COOH + KOH → ONLY₃COOK + FAMILY₂O

f) (RCOO)₃OLD₃H₅ + 3NaOH \(\overset{t^{0}}{\rightarrow}\) C₃H₅(OH)₃ + 3RCOONa

g) ONLY₃COOC₂H₅ + FRIENDS₂O \(\xrightarrow[axit]{t^{0}}\) ONLY₃COOH + C₂H₅OH

h)₂H₄(k) + Surname₂O(l) \(\overset{Acid}{\rightarrow}\) C₂H₅OH

i) 2 ONLY₃COOH + Zn → (ONLY₃COO)₂Zn + HEAR₂\(\uparrow\)

k) 2 ONLY₃COOH + CaCO₃ → (ONLY₃COO)₂Ca + CO₂\(\uparrow\) + Surname₂O

2.2. Form 2: Properties of acetic acid

For 50 ml of acetic acid solution to completely react with Mg to evaporate the solution, we get 1.42 g of salt.

a. Calculate the molar concentration of the acidic solution.

b. Volume of gas H₂ how much was born in dktc?

Solution guide

Chemical equation:

2 ONLY₃COOH + Mg → (ONLY₃COO)₂Mg + H₂\(\uparrow\)

0.02 \(\leftarrow\) 0.01 → 0.01

The number of moles of salt formed is: \({n_{{{(C{H_3}COO)}_2}Mg}} = \frac{m}{M} = \frac{{1,42}}{{142} } = 0.01(mol)\)

a) The molar concentration of the acidic solution is:

\({C_M} = \frac{n}{V} = \frac{{0.02}}{{0.005}} = 0.4(M)\)

b) Volume of gas H₂ born:

\({V_{{H_2}}} = 0.01,22.4 = 0.224(lit)\)

2.3. Form 3: Esterification reaction

There is a mixture A of ethyl alcohol and acetic acid. The experiment was carried out with mixture A and obtained the following results:
– If A reacts with excess sodium, after the reaction, 4.48 liters of colorless gas are obtained.
– If A reacts with Na₂HAVE₃ excess and pass the gas formed through the flask of lime water in the excess, 10g of precipitate is obtained.
a) Write chemical equations.
b) Calculate the mass percent of each substance in mixture A.
The gas volumes were all measured under standard conditions.

Solution guide

Let a and b be the number of moles of alcohol and acid, respectively
Reaction Equation:

2 C₂H₅OH + 2Na → 2C₂H₅ONa + FAMILY₂

a (mol) → \(\frac{a}{2}\)(mol)

2 ONLY₃COOH + 2Na → 2CH₃COONa + FAMILY₂

b (mol) → \(\frac{b}{2}\) (mol)

Because when A reacts with excess sodium, after the reaction, 4.48l of colorless gas is obtained, we have:

\({n_{{H_2}}} = \frac{{(a + b)}}{2} = \frac{{4,48}}{{22.4}} = 0.2 \Rightarrow a + b = 0.4\) (1)

When acting with Na₂HAVE₃ Only acids can produce CO .₂

Reaction Equation:

2 ONLY₃COOH + Na₂HAVE₃ → (ONLY₃COO)₂Ca + CO₂ + FRIENDS₂O

0.2 \(\leftarrow\) 0.1

HAVE₂ + Ca(OH)₂ (excess) → CaCO₃\(\downarrow\) + Surname₂O

0.1 \(\leftarrow\) 0.1

Number of moles of CaCO . precipitate₃ is: \({n_{CaC{O_3}}} = \frac{{10}}{{100}} = 0.1(mol)\)

So the number of moles of acid is 0.2 (or b = 0.2) a = 0.2 (mol)

The mass of mixture A is:

m = m_{acetic acid} + m_{Ethyl alcohol} = 0.2.46 + 0.2.60 = 21.2 (grams)

The mass percent of ethyl alcohol is:

\(\% {m_{{C_2}{H_5}OH}} = \frac{{0,2.46}}{{21,2}}.100 = 43.4\)

The mass percent of acetic acid is:

\(\% {m_{C{H_3}COOH}} = 100 – \% {m_{{C_2}{H_5}OH}} = 100 – 43.4 = 56.6\)

2.4. Saponification reaction

A type of fat prepared from C₁₅H_{thirty first}COOH and glycerol C₃H₅(OH)₃

a) Write the equation for the reaction

b) Heat the above 4.03 kg of fat with excess amount of NaOH solution. Calculate the mass of glycerol obtained.

c) Calculate the mass of bar soap containing 72% salt C₁₅H_{thirty first}COONa is modifiable.

Solution guide

a) Chemical equation:

3C₁₅H_{thirty first}COOH + C₃H₅(OH)₃ → (C₁₅H_{thirty first}COO)₃OLD₃H₅ + 3 HOURS₂O

b) The number of moles of fat reacting is:

\({n_{{{({C_{15}}{H_{31}}COO)}_3}{C_3}{H_5}}} = \frac{{4,03}}{{8,06}} = 0.5(kmol)\)

(C)₁₅H_{thirty first}COO)₃OLD₃H₅ + 3NaOH (excess) \(\overset{t^{0}}{\rightarrow}\) 3C₁₅H_{thirty first}COONa + OLD₃H₅(OH)₃

0.5 (kmol) → 1.5 (kmol) 0.5 (kmol)

The mass of glycerol obtained is:

\({m_{{C_3}{H_5}{{\left( {OH} \right)}_3}}} = 0.5.92 = 46(kg)\)

c) Mass of bar soap containing 72% salt C₁₅H_{thirty first}The modulated COONa is:

m_Soap \(= 1.5.267.\frac{{100}}{{72}} = 556.25(kg)\)

3. Practice

3.1. Essay exercises

Question 1: Heat 6g ONLY₃COOH with 9.2 grams of C₂H₅OH (with H₂SO₄ catalyst) until the reaction reaches equilibrium, 5.5 grams of ester are obtained. Calculate the efficiency of the esterification reaction.

Verse 2: Mixture X includes HCOOH and CH₃COOH (molar ratio 1:1). Take 5.3 grams of X to work with 5.75 grams of C₂H₅OH (catalyst H₂SO₄ concentrated) to obtain mg of the ester mixture (the efficiency of the esterification reactions is 80%). Calculate the value of m.
Question 3: For 0.3 mol of monomeric acid X mixed with 0.25 mol of ethyl alcohol to carry out esterification reaction to obtain 18 grams of ester. Separate the excess alcohol and acid to react with Na and see the release of 2.128 liters of H₂. Determine the acid formula and the esterification reaction yield.

Question 4: Let 4 moles of acetic acid react with a mixture of 0.5 moles of glycerol and 1 mole of etilenglicol (catalyst H₂SO₄). Calculate the mass of product obtained out of water given that 50% acid and 80% each alcohol react.

3.2. Multiple choice exercises

Question 1: The substance that releases the most energy when oxidizing food is:

A. Protein

B. Powder

C. Fat

D. Fiber

Verse 2: Which of the following is a formula for fat?

A. R-COOH

B. OLD₁₇H₃₅-COOH

C. OLD₃H₅(OH)₃

D. (C₁₇H₃₅-COO)₃OLD₃H₅

Question 3: There are three unlabelled vials: ethyl alcohol, acetic acid, cooking oil. How can the following be distinguished?

A. Use red litmus and water.

B. Carbon dioxide gas and water.

C. Sodium metal and water.

D. Phenolphthalein and water.

4. Conclusion

After the lesson, you should know the following:

– Structural formula, structural features, chemical properties (characteristic reactions) and relationships between substances.

– Preparation and main application of ethyl alcohol, acetic acid and fat.

– Determine the correct molecular formula, structural formula when knowing the properties of the substance.