Question

Вычислите массу осадка М, Н полученного при взаимодействии раствора содержащего 0,1моль сульфата меди, с избытком раствора KOH

Answer 1

    0,1 моль                                 Х г

   CuSO4      +      2KOH =      Cu(OH)2      +     K2SO4

n = 1 моль                             n = 1 моль

                                              Mr = 97,5 г/моль

                                              m = 97,5 г

0,1 моль CuSO4     -    Х г Cu(OH)2

1 моль  CuSO4       -    97,5 г Cu(OH)2

m(Cu(OH)2) = 0,1 * 97,5 / 1 = 9,75 г

Answer 2

Calculation of the Mass of Precipitate

To calculate the mass of the precipitate (M, N) obtained during the interaction of a solution containing 0.1 moles of copper sulfate (CuSO4) with an excess of potassium hydroxide (KOH) solution, we need to use the balanced chemical equation for the reaction between copper sulfate and potassium hydroxide.

The balanced chemical equation for the reaction is as follows: CuSO4 + 2KOH -> Cu(OH)2 + K2SO4

From the balanced equation, we can see that 1 mole of copper sulfate (CuSO4) reacts with 2 moles of potassium hydroxide (KOH) to produce 1 mole of copper hydroxide (Cu(OH)2) and 1 mole of potassium sulfate (K2SO4).

Given that the solution contains 0.1 moles of copper sulfate (CuSO4), we can determine the number of moles of copper hydroxide (Cu(OH)2) formed by using the stoichiometry of the reaction.

According to source the molar mass of copper sulfate (CuSO4) is 159.6 g/mol.

Using this information, we can calculate the mass of the precipitate (M, N) as follows:

1. Calculate the number of moles of copper sulfate (CuSO4): Number of moles = 0.1 moles

2. Calculate the number of moles of copper hydroxide (Cu(OH)2) formed: According to the balanced equation, 1 mole of copper sulfate (CuSO4) reacts to form 1 mole of copper hydroxide (Cu(OH)2). Therefore, the number of moles of copper hydroxide (Cu(OH)2) formed is also 0.1 moles.

3. Calculate the molar mass of copper hydroxide (Cu(OH)2): According to source the molar mass of copper hydroxide (Cu(OH)2) is 97.56 g/mol.

4. Calculate the mass of the precipitate (M, N): Mass of precipitate = Number of moles of copper hydroxide (Cu(OH)2) * Molar mass of copper hydroxide (Cu(OH)2) Mass of precipitate = 0.1 moles * 97.56 g/mol

Using the above calculations, we can determine the mass of the precipitate (M, N) obtained during the interaction of the solution containing 0.1 moles of copper sulfate (CuSO4) with an excess of potassium hydroxide (KOH) solution.

Please note that the exact value of the mass of the precipitate cannot be determined without knowing the volume of the solution. The calculations provided above only give the mass of the precipitate per mole of copper sulfate reacted.

The mass of the precipitate (M, N) obtained during the interaction of the solution containing 0.1 moles of copper sulfate (CuSO4) with an excess of potassium hydroxide (KOH) solution is approximately 9.756 grams.