Question

Какую массу фосфата кальция [Ca3(PO4)2] можно получить при добавлении избытка ортофосфорной кислоты (H3PO4) к 150 мл 10% раствора хлорида кальция [CaCl2]%3F Схема химической реакции%3A CaCl2 %2B H3PO4  Ca3(PO4)2 %2BHCl

Answer 1

3 CaCl₂ + 2 H₃PO₄ = Ca₃(PO₄)₂↓ + 6 HCl

Плотность 10% раствора CaCl₂  - 1,0835 г/мл (табличные данные)
Масса 150 мл этого раствора равна 150 мл *  1,0835 г/мл   = 162,525 г.
10% этой массы - масса чистого СaCl₂: m(CaCl₂) = 16.2525 г.

M(CaCl₂) = 40+2*35,45 = 110,9 г/моль
M(Ca₃(PO₄)₂) = 310,2 г/моль 

3 CaCl₂     +    2 H₃PO₄ = Ca₃(PO₄)₂↓ + 6 HCl
3*110,9 г                            310,2 г
16,2525 г                              х г.

х = 16,2525*310,2 / (3*110,9) = 15,15 ≈ 15 г.
Ответ: 15 г.

Answer 2

Calculation of the Mass of Calcium Phosphate (Ca3(PO4)2)

To calculate the mass of calcium phosphate (Ca3(PO4)2) that can be obtained when an excess of orthophosphoric acid (H3PO4) is added to a 150 ml 10% solution of calcium chloride (CaCl2), we need to consider the balanced chemical equation for the reaction:

CaCl2 + H3PO4 → Ca3(PO4)2 + HCl

From the equation, we can see that 1 mole of calcium chloride (CaCl2) reacts with 2 moles of orthophosphoric acid (H3PO4) to produce 1 mole of calcium phosphate (Ca3(PO4)2).

To determine the amount of calcium chloride in moles, we need to convert the volume of the 10% solution to moles.

The molar mass of calcium chloride (CaCl2) is 110.98 g/mol, and the molar mass of orthophosphoric acid (H3PO4) is 97.99 g/mol.

Let's calculate the amount of calcium chloride in moles:

Step 1: Calculate the mass of calcium chloride in the 150 ml 10% solution.

The concentration of the calcium chloride solution is 10%, which means there are 10 grams of calcium chloride in 100 ml of solution.

Mass of calcium chloride in 150 ml solution = (10 g/100 ml) * 150 ml = 15 g

Step 2: Convert the mass of calcium chloride to moles.

Moles of calcium chloride = (mass of calcium chloride)/(molar mass of calcium chloride)

Moles of calcium chloride = 15 g / 110.98 g/mol = 0.135 moles

Step 3: Calculate the moles of calcium phosphate (Ca3(PO4)2) that can be obtained.

From the balanced chemical equation, we know that 1 mole of calcium chloride reacts with 1 mole of calcium phosphate.

Therefore, the moles of calcium phosphate = 0.135 moles

Step 4: Calculate the mass of calcium phosphate (Ca3(PO4)2).

Mass of calcium phosphate = (moles of calcium phosphate) * (molar mass of calcium phosphate)

The molar mass of calcium phosphate (Ca3(PO4)2) is 310.18 g/mol.

Mass of calcium phosphate = 0.135 moles * 310.18 g/mol = 41.92 g

Therefore, the mass of calcium phosphate (Ca3(PO4)2) that can be obtained is approximately 41.92 grams.

Please note that the above calculation assumes ideal conditions and complete reaction conversion. In practice, the actual yield may be lower due to various factors such as side reactions, impurities, and incomplete conversion.

Chemical Reaction:

The chemical reaction between calcium chloride (CaCl2) and orthophosphoric acid (H3PO4) to form calcium phosphate (Ca3(PO4)2) and hydrochloric acid (HCl) can be represented as follows:

CaCl2 + H3PO4 → Ca3(PO4)2 + HCl

Please let me know if there is anything else I can help you with!