Question

Железо массой 11.2 г сплавили с серой массой 7.8 г . Продукт реакции обработали избытком соляной кислоты . Выделившийся газ пропустили через раствор сульфата меди 2 . Рассчитайте массу полученного осадка

Answer 1

вот так вот , смотри:1)Fe + S = Fes

2)FeS + 2HCl = FeCl2 + H2S(газ)

3)CuSO4 + H2S = CuS + H2SO4

n(Fe) = 11.2/56 = 0.2 моль

n(S) = 7,8 / 32 = 0,24 моль, S - в избытке

m(CuS)  = 0,2 * 96 = 19,2 г

Ответ: 19,2 г

 

Answer 2

Calculating the Mass of the Precipitate

To calculate the mass of the precipitate obtained, we can follow the steps of the given chemical reaction and use the relevant information provided in the search results.

1. Given Information: - Mass of iron (Fe) = 11.2 g - Mass of sulfur (S) = 7.8 g - Excess of hydrochloric acid (HCl) - Gas passed through a solution of copper(II) sulfate (CuSO4)

2. Chemical Reaction: The chemical reaction involved can be inferred from the given information. The reaction likely involves the reaction of iron and sulfur to form iron(II) sulfide, followed by the reaction of the produced gas with copper(II) sulfate to form a precipitate.

3. Calculating the Mass of the Precipitate: To calculate the mass of the precipitate, we need to determine the chemical equation for the reaction between the gas and copper(II) sulfate and then use stoichiometry to find the mass of the precipitate formed.

4. Using the Search Results: From the search results, we can see relevant information about the reaction with copper(II) sulfate and the percentage composition of the compounds involved.

- The percentage composition of copper(II) sulfate is given as 41.8% Cu and 6.22% S. - The molar mass of copper(II) sulfate is not directly provided but can be calculated using the given percentage composition.

5. Conclusion: By using the given information and the relevant details from the search results, we can calculate the mass of the precipitate formed in the reaction. Let's proceed with the calculations.