Question

Только главный мозг или профессор помогите пожалуйста очень важно 1) До розчину масою 95 г із масовою часткою магній хлориду 0,1 долили розчин натрій гідроксиду до повної взаємодії. Обчисліть масу утвореного магній гідроксиду. (4,66 г) 2)До розчину лугу масою 34,2 г з масовою часткою барій гідроксиду 0,1 долили розчин сульфатної кислоти до повної нейтралізації. Обчисліть масу утвореного осаду. 3)До розчину масою 34 г з масовою часткою арґентум(І) нітрату 0,3 долили розчин натрій хлориду до повної їх взаємодії. Обчисліть кількість речовини утвореного арґентум(І) хлориду, що випав в осад.(0,06 моль)

Answer 1

Ответ:

Объяснение:

1)

Дано:

m(р-нMgCl₂) = 95 г

w(MgCl₂) =  0,1

Найти: m(Mg(OH)₂) - ?  

Знайдемо масу MgCl₂ в розчині:

m(MgCl₂)  = m(р-нMgCl₂) * w(MgCl₂) = 95 * 0,1 = 9,5 г

Складемо РХР:

9,5 г                          х г

MgCl₂ + 2 NaOH = Mg(OH)₂↓ + 2 NaCl

n = 1 моль              n = 1 моль

M = 95 г/моль       М = 58 г/моль                          

m = 95 г                 m =  58 г

m(Mg(OH)₂) = 9,5 * 58 : 95 = 5,8 г

2)

Дано:

m(р-нBa(OH)₂) = 34,2 г

w(Ba(OH)₂) =  0,1

Найти: m(BaSO₄) - ?  

Знайдемо масу Ba(OH)₂ в розчині:

m(Ba(OH)₂)  = m(р-нBa(OH)₂) * w(Ba(OH)₂) = 34,2 * 0,1 = 3,42 г

Складемо РХР:

3,42 г                         х г

Ba(OH)₂ + H₂SO₄ = BaSO₄↓ + 2H₂O

n = 1 моль               n = 1 моль

M = 171 г/моль        М = 233 г/моль                          

m = 171 г                  m =  233 г

m(BaSO₄) = 3,42 * 233 : 171 = 4,66 г

3)

Дано:

m(р-нAgNO₃) = 34 г

w(AgNO₃) =  0,3

Найти: n(AgCl) - ?  

Знайдемо масу AgNO₃ в розчині:

m(AgNO₃)  = m(р-нAgNO₃) * w(AgNO₃) = 34 * 0,3 = 10,2 г

Складемо РХР:

NaCl + AgNO₃ = AgCl + NaNO₃

М(AgNO₃) = 108 + 14 + 16 * 3 = 170 г/моль

n(AgNO₃) = m(AgNO₃)/М(AgNO₃) = 10,2/170 = 0,06 моль

З РХР бачимо, що n(AgNO₃) : n(AgCl) = 1 : 1, тобто

n(AgCl) = n(AgNO₃) = 0,06 моль          

Answer 2

Calculation of Magnesium Hydroxide Formation

To calculate the mass of magnesium hydroxide formed, we need to determine the number of moles of magnesium chloride and sodium hydroxide reacted and then use stoichiometry to find the number of moles of magnesium hydroxide formed.

Given: - Mass of magnesium chloride solution: 95 g - Mass fraction of magnesium chloride: 0.1 - Mass of sodium hydroxide solution added: unknown

To find the mass of sodium hydroxide solution added, we can use the mass fraction of magnesium chloride. The mass of magnesium chloride can be calculated as follows:

Mass of magnesium chloride = Mass of solution x Mass fraction of magnesium chloride Mass of magnesium chloride = 95 g x 0.1 = 9.5 g

Since the mass of magnesium chloride is given, we can now calculate the number of moles of magnesium chloride using its molar mass. The molar mass of magnesium chloride (MgCl2) is 95.211 g/mol.

Number of moles of magnesium chloride = Mass of magnesium chloride / Molar mass of magnesium chloride Number of moles of magnesium chloride = 9.5 g / 95.211 g/mol = 0.0998 mol

According to the balanced chemical equation for the reaction between magnesium chloride and sodium hydroxide, 1 mole of magnesium chloride reacts with 2 moles of sodium hydroxide to form 1 mole of magnesium hydroxide.

Therefore, the number of moles of sodium hydroxide required is twice the number of moles of magnesium chloride:

Number of moles of sodium hydroxide = 2 x Number of moles of magnesium chloride Number of moles of sodium hydroxide = 2 x 0.0998 mol = 0.1996 mol

To find the mass of sodium hydroxide solution added, we need to know the concentration of the sodium hydroxide solution. Without this information, we cannot calculate the mass of the formed magnesium hydroxide.

Unfortunately, the given information does not provide the concentration or volume of the sodium hydroxide solution added. Therefore, we cannot calculate the mass of the formed magnesium hydroxide.

Calculation of Barium Sulfate Precipitate

To calculate the mass of the barium sulfate precipitate formed, we need to determine the number of moles of barium hydroxide and sulfuric acid reacted and then use stoichiometry to find the number of moles of barium sulfate formed.

Given: - Mass of barium hydroxide solution: 34.2 g - Mass fraction of barium hydroxide: 0.1 - Mass of sulfuric acid solution added: unknown

Using the mass fraction of barium hydroxide, we can calculate the mass of barium hydroxide:

Mass of barium hydroxide = Mass of solution x Mass fraction of barium hydroxide Mass of barium hydroxide = 34.2 g x 0.1 = 3.42 g

Since the mass of barium hydroxide is given, we can now calculate the number of moles of barium hydroxide using its molar mass. The molar mass of barium hydroxide (Ba(OH)2) is 171.34 g/mol.

Number of moles of barium hydroxide = Mass of barium hydroxide / Molar mass of barium hydroxide Number of moles of barium hydroxide = 3.42 g / 171.34 g/mol = 0.01998 mol

According to the balanced chemical equation for the reaction between barium hydroxide and sulfuric acid, 1 mole of barium hydroxide reacts with 1 mole of sulfuric acid to form 1 mole of barium sulfate.

Therefore, the number of moles of sulfuric acid required is equal to the number of moles of barium hydroxide:

Number of moles of sulfuric acid = Number of moles of barium hydroxide = 0.01998 mol

To find the mass of sulfuric acid solution added, we need to know the concentration of the sulfuric acid solution. Without this information, we cannot calculate the mass of the formed barium sulfate.

Unfortunately, the given information does not provide the concentration or volume of the sulfuric acid solution added. Therefore, we cannot calculate the mass of the formed barium sulfate.

Calculation of Silver Chloride Precipitate

To calculate the amount of silver chloride precipitate formed, we need to determine the amount of silver nitrate and sodium chloride reacted and then use stoichiometry to find the amount of silver chloride formed.

Given: - Mass of silver nitrate solution: 34 g - Mass fraction of silver nitrate: 0.3 - Amount of sodium chloride solution added: unknown

Using the mass fraction of silver nitrate, we can calculate the mass of silver nitrate:

Mass of silver nitrate = Mass of solution x Mass fraction of silver nitrate Mass of silver nitrate = 34 g x 0.3 = 10.2 g

Since the mass of silver nitrate is given, we can now calculate the number of moles of silver nitrate using its molar mass. The molar mass of silver nitrate (AgNO3) is 169.87 g/mol.

Number of moles of silver nitrate = Mass of silver nitrate / Molar mass of silver nitrate Number of moles of silver nitrate = 10.2 g / 169.87 g/mol = 0.060 mol

According to the balanced chemical equation for the reaction between silver nitrate and sodium chloride, 1 mole of silver nitrate reacts with 1 mole of sodium chloride to form 1 mole of silver chloride.

Therefore, the number of moles of sodium chloride required is equal to the number of moles of silver nitrate:

Number of moles of sodium chloride = Number of moles of silver nitrate = 0.060 mol

To find the amount of sodium chloride solution added, we need to know the concentration of the sodium chloride solution. Without this information, we cannot calculate the amount of the formed silver chloride.

Unfortunately, the given information does not provide the concentration or volume of the sodium chloride solution added. Therefore, we cannot calculate the amount of the formed silver chloride.