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 = macetic acid + mEthyl 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 C15Hthirty firstCOOH 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 C15Hthirty firstCOONa is modifiable.
Solution guide
a) Chemical equation:
3C15Hthirty firstCOOH + C3H5(OH)3 → (C15Hthirty firstCOO)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)15Hthirty firstCOO)3OLD3H5 + 3NaOH (excess) \(\overset{t^{0}}{\rightarrow}\) 3C15Hthirty firstCOONa + 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 C15Hthirty firstThe modulated COONa is:
mSoap \(= 1.5.267.\frac{{100}}{{72}} = 556.25(kg)\)
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.
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