Question

К 80 Г раствора с массовой долей гидроксида натрия 5% добавили избыток раствора сульфата меди(2). Определите массу выпавшего осадка. Помогите пожалуйста, срочно нужно❤️

Answer 1

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Answer 2

Calculating the Mass of Precipitate

To calculate the mass of the precipitate formed when an excess of copper(II) sulfate solution is added to an 80g solution with a 5% mass fraction of sodium hydroxide, we can use the concept of chemical reactions and stoichiometry.

The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and copper(II) sulfate (CuSO4) is:

CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4

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

First, we need to calculate the amount of sodium hydroxide in the 80g solution. Then, we can determine the limiting reactant and use it to find the mass of the precipitate formed.

Given: - Mass fraction of sodium hydroxide in the solution = 5% - Total mass of the solution = 80g

Calculating the Amount of Sodium Hydroxide

The mass of sodium hydroxide in the solution can be calculated using the mass fraction.

Mass of NaOH = Mass fraction × Total mass of solution Mass of NaOH = 5% × 80g = 0.05 × 80g = 4g

Determining the Limiting Reactant

Next, we need to determine the limiting reactant by comparing the amount of sodium hydroxide with the amount required to react with the copper(II) sulfate.

The molar mass of NaOH is approximately 40 g/mol, and the molar mass of CuSO4 is approximately 160 g/mol.

Number of moles of NaOH = Mass of NaOH / Molar mass of NaOH Number of moles of NaOH = 4g / 40 g/mol = 0.1 mol

Since the balanced chemical equation shows that 1 mole of CuSO4 reacts with 2 moles of NaOH, we can calculate the amount of CuSO4 required for complete reaction with the NaOH.

Number of moles of CuSO4 required = 0.1 mol / 2 = 0.05 mol

Calculating the Mass of Precipitate

Now that we have determined the limiting reactant, we can calculate the mass of the precipitate formed, which is copper(II) hydroxide (Cu(OH)2).

The molar mass of Cu(OH)2 is approximately 97 g/mol.

Mass of Cu(OH)2 = Number of moles of Cu(OH)2 × Molar mass of Cu(OH)2 Mass of Cu(OH)2 = 0.05 mol × 97 g/mol = 4.85g

Therefore, the mass of the precipitate formed when an excess of copper(II) sulfate solution is added to the 80g solution with a 5% mass fraction of sodium hydroxide is approximately 4.85 grams.