Question

Обчислити масу осаду, що утвориться при змішуванні 20 г розчину ферум (ІІ) сульфату з масовою часткою розчиненої речовини 8% з розчином натрій гідроксиду взятим в достатній кількості. Срочно!!!Даю 80 балів​

Answer 1

Объяснение:

m(FeSO4)=m•w= 20•0.08=1.6г

FeSO4 Молярная масса: 152 г/моль

Fe(OH)₂ Молярная масса: 90 г/моль

1.6г. хг.

2NaOH + FeSO4 = Na2SO4 + Fe(OH)2↓

152г/моль 90г/моль

х=(1.6•90)/152= 0.95г гидроксида железа (|||)

Answer 2

Calculation of the Mass of Precipitate Formed

To calculate the mass of the precipitate formed when 20 g of a solution of iron(II) sulfate with a mass fraction of dissolved substance of 8% is mixed with a sufficient amount of sodium hydroxide solution, we need to consider the reaction between iron(II) sulfate and sodium hydroxide.

The balanced chemical equation for the reaction is as follows:

FeSO4 + 2NaOH -> Fe(OH)2 + Na2SO4

From the balanced equation, we can see that 1 mole of iron(II) sulfate reacts with 2 moles of sodium hydroxide to form 1 mole of iron(II) hydroxide and 1 mole of sodium sulfate.

To calculate the mass of the precipitate formed, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

Let's calculate the number of moles of iron(II) sulfate and sodium hydroxide in the given quantities:

- Mass of iron(II) sulfate solution = 20 g - Mass fraction of dissolved substance in iron(II) sulfate solution = 8% - Mass of dissolved substance in iron(II) sulfate solution = (8/100) * 20 g = 1.6 g

Using the molar mass of iron(II) sulfate (FeSO4) and sodium hydroxide (NaOH), we can calculate the number of moles:

- Molar mass of FeSO4 = 55.85 g/mol + 32.07 g/mol + (4 * 16.00 g/mol) = 151.91 g/mol - Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol

Number of moles of iron(II) sulfate = 1.6 g / 151.91 g/mol Number of moles of sodium hydroxide = (mass of sodium hydroxide solution) / (molar mass of NaOH)

Since the mass of sodium hydroxide solution is not provided, we cannot calculate the exact number of moles of sodium hydroxide. However, we can assume that a sufficient amount of sodium hydroxide is added, which means that sodium hydroxide is in excess and iron(II) sulfate is the limiting reactant.

Therefore, the number of moles of iron(II) hydroxide formed is equal to the number of moles of iron(II) sulfate.

To calculate the mass of iron(II) hydroxide precipitate formed, we can use the molar mass of Fe(OH)2:

- Molar mass of Fe(OH)2 = 55.85 g/mol + (2 * 16.00 g/mol) + (2 * 1.01 g/mol) = 89.91 g/mol

Mass of iron(II) hydroxide precipitate formed = Number of moles of iron(II) hydroxide * Molar mass of Fe(OH)2

Now, let's calculate the mass of the precipitate formed:

Mass of iron(II) hydroxide precipitate formed = (Number of moles of iron(II) sulfate) * (Molar mass of Fe(OH)2)

Note: Since the mass of sodium hydroxide solution is not provided, we cannot calculate the exact mass of the precipitate formed. However, we can calculate the number of moles of iron(II) hydroxide formed, which will allow us to determine the molar mass and the mass of the precipitate formed.

Please provide the mass of the sodium hydroxide solution so that we can calculate the mass of the precipitate formed accurately.