Question

Можете пожалуйста помочь с химией.Отмечу как лучший ответ. Оксид фосфора(V) массой 14,2 г растворили в 180 г воды с образованием ортофосфорной кислоты. Затем в полученный раствор добавили 12 г твердого гидроксида натрия. Определите массовую долю солей в конечном растворе. В ответе запишите уравнения реакций, которые указаны в условии задачи, и приведите все необходимые вычисления (указывайте единицы измерения искомых физических величин).

Answer 1

Ответ:

Все еще нужно решение?

Объяснение:

Answer 2

Task Analysis

To solve this problem, we need to determine the mass fraction of salts in the final solution after reacting phosphorus pentoxide (P2O5) with water to form orthophosphoric acid (H3PO4), and then adding solid sodium hydroxide (NaOH) to the resulting solution. We are given the masses of P2O5, water, and NaOH. We need to write the balanced chemical equations for the reactions involved and perform the necessary calculations to find the mass fraction of salts in the final solution.

Given Information

- Mass of P2O5: 14.2 g - Mass of water: 180 g - Mass of NaOH: 12 g

Balanced Chemical Equations

1. Reaction between P2O5 and water to form H3PO4: ``` P2O5 + 3H2O -> 2H3PO4 ```

2. Reaction between H3PO4 and NaOH to form salts: ``` 2H3PO4 + 6NaOH -> 2Na3PO4 + 6H2O ```

Calculation Steps

1. Calculate the molar mass of P2O5, H3PO4, and NaOH. 2. Convert the given masses of P2O5, water, and NaOH to moles using their respective molar masses. 3. Determine the limiting reactant by comparing the moles of P2O5 and NaOH. 4. Calculate the moles of H3PO4 and Na3PO4 formed based on the limiting reactant. 5. Calculate the mass of Na3PO4 formed using its molar mass and the moles of Na3PO4. 6. Calculate the mass fraction of salts in the final solution by dividing the mass of Na3PO4 by the total mass of the final solution (water + Na3PO4).

Let's perform the calculations step by step.

Step 1: Calculate the Molar Masses

The molar masses of the compounds involved are: - P2O5: 141.94 g/mol - H3PO4: 97.99 g/mol - NaOH: 39.997 g/mol

Step 2: Convert Masses to Moles

- Moles of P2O5: 14.2 g / 141.94 g/mol = 0.100 mol - Moles of water: 180 g / 18.015 g/mol = 9.99 mol - Moles of NaOH: 12 g / 39.997 g/mol = 0.300 mol

Step 3: Determine the Limiting Reactant

To determine the limiting reactant, we compare the moles of P2O5 and NaOH. The balanced equation shows that the stoichiometric ratio between P2O5 and NaOH is 1:6. Therefore, 1 mole of P2O5 reacts with 6 moles of NaOH.

The moles of NaOH available (0.300 mol) are more than enough to react with the moles of P2O5 (0.100 mol). Therefore, P2O5 is the limiting reactant.

Step 4: Calculate Moles of H3PO4 and Na3PO4

From the balanced equation, we can see that 1 mole of P2O5 reacts to form 2 moles of H3PO4. Therefore, the moles of H3PO4 formed are 2 times the moles of P2O5.

- Moles of H3PO4: 2 * 0.100 mol = 0.200 mol

From the balanced equation, we can also see that 2 moles of H3PO4 react to form 2 moles of Na3PO4. Therefore, the moles of Na3PO4 formed are the same as the moles of H3PO4.

- Moles of Na3PO4: 0.200 mol

Step 5: Calculate the Mass of Na3PO4

The molar mass of Na3PO4 is 163.94 g/mol.

- Mass of Na3PO4: 0.200 mol * 163.94 g/mol = 32.79 g

Step 6: Calculate the Mass Fraction of Salts

The total mass of the final solution is the sum of the masses of water and Na3PO4.

- Total mass of the final solution: 180 g (water) + 32.79 g (Na3PO4) = 212.79 g

The mass fraction of salts in the final solution is the mass of Na3PO4 divided by the total mass of the final solution.

- Mass fraction of salts: 32.79 g / 212.79 g = 0.154 (rounded to three decimal places)

Answer

The mass fraction of salts in the final solution is approximately 0.154.

The balanced chemical equations for the reactions involved are: 1. P2O5 + 3H2O -> 2H3PO4 2. 2H3PO4 + 6NaOH -> 2Na3PO4 + 6H2O

Please let me know if you need any further assistance!