Giải bài 1 2 3 4 5 6 7 8 9 trang 118 119 sgk Hóa Học 12

Instructions for solving Lesson 26. Alkaline earth metals and their important compounds in Chemistry 12 textbook. Contents Solution 1 2 3 4 5 6 7 8 9 pages 118 119 Chemistry textbook 12 fully included theory and exercises, with formulas, chemical equations, topics, etc., are included in the textbooks to help students study well in chemistry 12, prepare for the national high school graduation exam.

THEORY

A. Alkaline Earth Metals

1. Location and structure

Alkaline earth metals are s(ns2) elements of group IIA, including metals:

Berium, Magnesium, Calcium, Strontium, Barium.

In each period, the alkaline earth metals come after the alkali metals.

2. Physical properties

Relatively low melting and boiling points.

– Due to the crystal lattice structure of different elements, the melting and boiling points do not change depending on the nuclear charge.

(Be, Mg has a hexagonal lattice; Ca, Sr has a face-centered cubic lattice; Ba has a mass-centered cubic lattice).

3. Chemical properties

Alkali earth metals have 2 outermost electrons in the e . configuration

⇒ tends to give up 2 electrons when participating in chemical reactions

⇒ Alkali earth metals have strong reducing properties.

a) Reaction with nonmetals

2Mg + O2 (xrightarrow{{{t}^{o}}}) 2MgO

Ca + Cl2 (xrightarrow{{{t}^{o}}}) CaCl2

b) React with acids

Ca + 2HCl → CaCl2 + H2

c) React with water

Ca + 2H2O → Ca(OH)2 + H2

Mg + H2O (xrightarrow{{{t}^{o}}}) MgO + H2

4. Application

– Be metal is used as an additive to make alloys with high elasticity, durability, and non-corrosion.

Mg metal is used to make alloys with hard, light and durable properties. These alloys are used to make airplanes, rockets, cars, etc. Mg metal is also used to synthesize many organic compounds. Mg powder mixed with oxidizing agent is used to make night light.

– Metal Ca is used as a reducing agent to separate oxygen and sulfur from steel. Calcium is also used to dry some organic compounds.

5. Modulation:

Electrolysis of molten alkaline earth metal salts.

CaCl2 (xrightarrow{text{pnc}}) Ca + Cl2

B. Some compounds of calcium

1. Calcium hydroxide, Ca(OH)2

Physical properties: It is a white solid, sparingly soluble in water.

– Chemical properties: Full of properties of an alkaline solution (reacts with acids, acid oxides, salts).

2. Calcium carbonate, CaCO3

Physical properties: white solid, insoluble in water.

Chemical properties: this is a salt of a weak acid, unstable, so it can react with many inorganic acids, releasing carbon dioxide:

CaCO3 + 2HCl → CaCl2 + H2O + CO2

CaCO3 + 2CH3COOH → Ca(CH3COO)2 + H2O + CO2

Calcium carbonate gradually dissolves in water containing carbon dioxide gas, forming the soluble salt calcium hydrocarbonate (Ca(HCO3)2):

CaCO3 + H2O + CO2 (rightleftarrows ) Ca(HCO3)2

⇒ Forward reaction: Explain the erosion of rainwater towards limestone.

Reverse reaction: Explain the formation of stalactites present in caves.

3. Calcium sulfate, CaSO4

– Calcium sulfate is a solid, white color, slightly soluble in water (solubility at 25oC is 0.15 g/100g H2O).

Depending on the amount of water of crystallization in calcium sulfate salts, we have 3 types:

+ CaSO4.2H2O found in nature is living gypsum, stable at normal temperature.

+ CaSO4.H2O or CaSO4.0.5H2O is calcined gypsum.

+ CaSO4 is called anhydrous gypsum.

– Anhydrous plaster is insoluble and does not react with water.

C. Hard water

– Definition: Hard water is water that contains many cations Ca2+, Mg2+. Water that contains little or no ions is called soft water.

– Classification:

+ Temporary hard water: is water containing ions: Ca2+, Mg2+, HCO3-

+ Permanent hard water: is water containing ions: Ca2+, Mg2+, SO42-, Cl-

Total Hard Water: Water that has both temporary and permanent hardness.

See also:: Instructions for solving problems 1 2 3 4 pages 41 Chemistry textbook 9

⇒ Natural water is usually completely hard water.

Harm of hard water:

+ Reduces foam, reduces the ability of soap to wash, makes food take longer to cook and reduces taste.

+ Hard water also harms manufacturing industries, damaging many solutions to be prepared.

– Measures to soften hard water: Principle: Reduce the concentration of cations Ca2+, Mg2+ in hard water.

Precipitation method:

• Temporarily hard water: Bring to a boil

Ca(HCO3)2 (xrightarrow{{{t}^{o}}}) CaCO3¯ + CO2 + H2O

Mg(HCO3)2 (xrightarrow{{{t}^{o}}}) MgCO3¯ + CO2 + H2O

• Permanent hard water: Use Na2CO3, Na3PO4 solution.

Ca2+ + (C{{O}_{3}}^{2-}) → CaCO3 ¯

3Ca2+ + (P{{O}_{4}}^{3-}) → Ca3(PO4)2 ¯

Ion exchange method.

– Identify Ca2+, Mg2+ ions in solution: To prove the presence of Ca2+ and Mg2+ ions, we use a solution containing carbonate salts to create CaCO3 or MgCO3 precipitates. Then aerate the excess CO2 into the solution, if the precipitate dissolves, it indicates the presence of Ca2+ or Mg2+ in the original solution.

EXERCISE

Below is the most complete and concise part of the Guide to Solving Lesson 1 2 3 4 5 6 7 8 9 pages 118 119 Chemistry 12 textbooks. The detailed content of the exercises you can see below:

1. Solve problem 1 page 118 turns 12

Arrange the alkaline earth metals in order of increasing nuclear charge, then

A. The atomic radius decreases gradually.

B. The ionization energy decreases gradually.

C. Decreasing reduction.

D. The ability to react with water decreases gradually.

See also:: Lesson 6 Page 119 Chemistry Textbook 10 – Song Thu Cable Television

Solution:

The alkaline earth metals are arranged in the order of increasing nuclear charge, then: atomic radius increases, ionization energy decreases, reducing capacity increases, ability to react with water increases.

So A, C, D are wrong. B is correct.

⇒ Answer: B.

2. Solve problem 2 page 119 turns 12

Adding Ca(OH)2 solution to Ca(HCO3)2 solution will

A. There is a white precipitate.

B. There are air bubbles coming out.

C. There is a white precipitate and air bubbles.

D. Nothing happened.

See also:: Lesson 6 Page 119 Chemistry Textbook 10 – Song Thu Cable Television

Solution:

We will see the appearance of a white precipitate of CaCO3 due to the reaction

$Ca(OH)_2 + Ca(HCO_3)_2 → 2CaCO_3↓ + 2H_2O$

⇒ Answer: A.

3. Solve problems 3 pages 119 turns 12

For 2.84 grams of a mixture of CaCO3 and MgCO3, it was completely absorbed with HCl solution, and 672 ml of CO2 was released (determined). The mass percent of the two salts (CaCO3, MgCO3) in the mixture is

A. 35.2% and 64.8%.

B. 70.4% and 29.6%.

C. 85.49% and 14.51%.

D. 17.6% and 82.4%.

See also:: Lesson 6 Page 119 Chemistry Textbook 10 – Song Thu Cable Television

Solution:

Let the number of moles of CaCO3 and MgCO3 be $x, y (mol)$ . respectively

Chemical equation:

$CaCO_3 + 2HCl → CaCl_2 + CO_2↑ + H_2O$

$x → x (mol)$

$MgCO_3 + 2HCl → MgCl_2 + CO_2↑ + H_2O$

$y → y (mol)$

According to the problem, we have the following system of equations:

(left{ begin{gathered} {m_{hh}} = 100x + 84y = 2.84 hfill {n_{C{O_2}}} = x + y = 0.03 hfill end{gathered} right. ⇒ left { begin{gathered} x = 0.02 hfill y = 0.01 hfill end{gathered} right.)

Therefore:

$⇒ %m_{CaCO_3} = frac{0.02.100}{2.84} .100% =70.4%$

$%m_{MgCO_3} = 100% – 70.4% = 29.6%$

⇒ Answer: B.

4. Solve problems 4 pages 119 turns 12

For 2 grams of a metal of group IIA, it is completely reacted with HCl solution to give 5.55 grams of chloride salt. Which of the following is a metal?

See also:: Solving problems 1,2,3, 4,5,6 page 67 Chemistry textbook 8: Switching between blocks

A. Be;

B. Mg;

C. Ca;

D. Ba.

See also:: Lesson 6 Page 119 Chemistry Textbook 10 – Song Thu Cable Television

Solution:

Method 1:

Let M be a group II metal, the chemical equation is:

$M + 2HCl → MCl_2 + H_2↑$

We have:

$m_{salt} = m_{KL} + m_{Cl^-}$

$⇒ n_{Cl^- , in ,salt} = dfrac{5.55-2}{35.5} = 0.1 (mol)$

$⇒n_M = dfrac{1}{2} .n_{Cl^-} = 0.05 (mol)$

(⇒ M = dfrac{2}{0.05} = 40 (g/mol))

⇒ The metal is $Ca$.

Method 2:

Let M be a group II metal, the number of moles is $x$

Chemical equation:

$M + 2HCl → MCl_2 + H_2↑$

According to the problem, we have a system of equations:

(left{ begin{array}{l}M{rm{x}} = 2x(M + 71) = 5.55end{array} right. to left{ begin{array}{l}x = 0.05 M = 40end{array} right.)

So M is $Ca$.

⇒ Answer: C.

5. Solve problems 5 pages 119 turns 12

For 2.8 grams of CaO to react with an excess of water, solution A is obtained. Boil 1.68 liters of CO2 (equivalent) into solution A.

a) Calculate the mass of the precipitate obtained.

b) What is the maximum mass of precipitate obtained when solution A is heated?

See also:: Lesson 6 Page 119 Chemistry Textbook 10 – Song Thu Cable Television

Solution:

a) Calculate the volume of the precipitate obtained

The number of moles of $CaO$ is:

({n_{CaO}} = frac{{2,8}}{{56}} = 0.05,,mol)

The number of moles of $CO_2$ is:

({n_{C{O_2}}} = frac{{1,68}}{{22.4}} = 0.075,,mol)

Chemical equation:

(CaO + {H_2}O to Ca{(OH)_2})

$0.05 → 0.05 (mol)$

(C{O_2} + Ca{(OH)_2} to CaC{{rm{O}}_3}↓ + {H_2}O)

$0.05 , , 0.05 → 0.05 (mol)$

We have:

({n_{CaC{O_3}}} = {n_{C{O_2}(pù)}} = {n_{Ca{{(OH)}_2}}} = 0.05,,mol))

({n_{C{O_2}(residual)}} = 0.075 – 0.05 = 0.025,,mol)

$CaCO_3$ forming $0.05 mol$ dissolved $0.025 mol$

(C{O_2} + CaC{{rm{O}}_3} + {H_2}O to Ca{(HC{O_3})_2})

$0.025 , , 0.025 → 0.025 (mol)$

Number of moles of $CaCO_3$ remaining:

${n_{CaC{O_3} ,remaining ,remaining}} = 0.05 – 0.025 = 0.025 mol$

The mass of precipitate obtained is:

${m_{CaC{O_3}}} = 0.025.100 = 2.5 grams$

b) When the solution is heated, the chemical equation is added:

(Ca(HCO_3)_2 overset{t^{o}}{rightarrow} CaCO_3↓ + CO_2↑ + H_2O)

(0.025 → 0.025 (mol))

So the maximum mass of precipitate obtained is:

${m_{CaC{O_3}}} = (0.025+0.025).100 = 5 grams$

6. Solve lesson 6 page 119 turns 12

When taking 14.25 grams of chloride salts of a metal with only valence II and a mass of nitrate salts of that metal having the same number of moles of chloride salts, a difference of 7.95 grams is found. Determine the name of metal.

See also:: Lesson 6 Page 119 Chemistry Textbook 10 – Song Thu Cable Television

Solution:

Call the metal with valence II to be searched for A.

⇒ The formula for chloride salt is $ACl_2$

And the nitrate salt formula is $A(NO_3)_2$

Set $n_{ACl_2} = n_{A(NO_3)_2} = x (mol)$

Method 1:

The mass difference is due to the origin ${NO_3}^-$ from the base $Cl^-$

Hence: $m_{A(NO_3)_2} – m_{ACl_2} = 7.95$

$⇒ (A + 124).x – (A + 71).x = 7.95$

$⇒ 53x = 7.95 x = 0.15 (mol)$

$⇒ M_{ACl_2} = dfrac{14,25}{0.15} = 95 (g/mol)$

$⇒ M_A = 95 – 71 = 24 (g/mol)$

⇒ Metal A is $Mg$.

Method 2:

According to the problem, we have a system of equations:

(left{ begin{array}{l}x(A + 71) = 14.25x(A + 124) = 14.25 + 7.95end{array} right. to left{ begin{array}{l} x = 0.15A = 24end{array} right.)

⇒ Metal A is $Mg$.

So the two salts are $MgCl_2$ and $Mg(NO_3)_2$.

7. Solve problems 7 page 119 turns 12

Dissolving 8.2 grams of CaCO3 and MgCO3 powder mixture in water requires 2,016 liters of CO2 (dktc). Determine the mass of each salt in the mixture.

See also:: Lesson 6 Page 119 Chemistry Textbook 10 – Song Thu Cable Television

Solution:

Let the number of moles of $CaCO_3$ and $MgCO_3$ in the mixture be $x, y (mol)$.

$CaCO_3 + CO_2 + H_2O → Ca(HCO_3)_2$

$x → x (mol)$

$MgCO_3 + CO_2 + H_2O → Mg(HCO_3)_2$

$y → y (mol)$

We have a system of equations:

(left{ begin{gathered} {m_{hh}} = 100x + 84y = 8.2 hfill {n_{C{O_2}}} = x + y = 0.09 hfill end{gathered} right. ⇒ left { begin{gathered} x = 0.04 hfill y = 0.05 hfill end{gathered} right.)

So we infer:

$m_{CaCO_3} = 0.04,100 = 4 (gram)$.

$m_{MgCO_3} = 84.0.05 = 4.2 (grams)$.

8. Solve problems 8 pages 119 turns 12

In a glass of water contains 0.01 mol Na+, 0.02 mol Ca2+, 0.01 mol Mg2+, 0.05 mol HCO3-, 0.02 mol Cl-. What type of water is in the cup?

A. Hard water has a temporary hardness.

B. Hard water has permanent hardness.

C. Hard water has total hardness.

D. Soft water.

See also:: Lesson 6 Page 119 Chemistry Textbook 10 – Song Thu Cable Television

Solution:

The upper cup of water contains ions:

$Ca^{2+}, Mg^{2+}, HCO3^-, Cl^-$

⇒ Belongs to full hard water.

⇒ Answer: C.

9. Solve lesson 9 page 119 turns 12

Write the chemical equation for the reaction to explain the use of Na3PO4 to soften hard water with total hardness.

See also:: Lesson 6 Page 119 Chemistry Textbook 10 – Song Thu Cable Television

Solution:

Chemical Equations:

$3Ca(HCO_3)_2 + 2Na_3PO_4 rightarrow Ca_3(PO_4)_2downarrow + 6NaHCO_3$

$3Mg(HCO_3)_2 + 2Na_3PO_4 rightarrow Mg_3(PO_4)_2downarrow + 6NaHCO_3$

$3CaCl_2 + 2Na_3PO_4 rightarrow Ca_3(PO_4)_2downarrow + 6NaCl$

$3CaSO_4 + 2Na_3PO_4 rightarrow Ca_3(PO_4)_2downarrow + 3Na_2SO_4$

Then all the ions $Ca^{2+}, Mg^{2+}$ are all precipitated as phosphate salts ⇒ softening the hard water completely.

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