Question

Помогите пожалуйста 20 баллов. Через 27 г раствора хлорида меди (II) пропустили избыток сероводорода. Определите массу выпавшего осадка, если выход составляет 84%​

Answer 1

Ответ:

На картинке ответ

Объяснение:

надеюсь помог

Answer 2

Calculation of the Mass of Precipitate

To calculate the mass of the precipitate formed when an excess of hydrogen sulfide (H2S) is passed through a solution of copper(II) chloride (CuCl2), we need to consider the balanced chemical equation for the reaction between copper(II) chloride and hydrogen sulfide:

CuCl2 + H2S -> CuS + 2HCl

From the balanced equation, we can see that one mole of copper(II) chloride reacts with one mole of hydrogen sulfide to produce one mole of copper(II) sulfide and two moles of hydrochloric acid.

To determine the mass of the precipitate, we need to know the molar mass of copper(II) sulfide (CuS). The molar mass of copper is approximately 63.55 g/mol, and the molar mass of sulfur is approximately 32.07 g/mol. Therefore, the molar mass of copper(II) sulfide is:

Molar mass of CuS = (1 * molar mass of Cu) + (1 * molar mass of S) = (1 * 63.55 g/mol) + (1 * 32.07 g/mol) = 95.62 g/mol

Next, we need to determine the number of moles of copper(II) sulfide formed. We can do this by using the stoichiometry of the balanced equation. Since the reaction has a yield of 84%, we can assume that 84% of the copper(II) chloride reacts to form copper(II) sulfide.

Let's assume we have x moles of copper(II) chloride. Since the reaction has a yield of 84%, the number of moles of copper(II) sulfide formed will be 84% of x moles.

Number of moles of CuS = 0.84 * x

Finally, we can calculate the mass of the precipitate using the molar mass of copper(II) sulfide:

Mass of CuS = Number of moles of CuS * Molar mass of CuS = (0.84 * x) * 95.62 g/mol

Please provide the initial number of moles of copper(II) chloride (x) in order to calculate the mass of the precipitate.