
What do you observe when ${H_2}S$ gas is passed through copper sulphate solution?
Answer
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Hint: We need to understand the reaction between ${H_2}S$ gas and copper sulfate solution and see what it produces. ${H_2}S$ is the chemical formula for hydrogen sulphide. It's a colourless chalcogen hydride gas that smells like rotting eggs. It's combustible, toxic, and corrosive. It is sometimes commonly known as hydrosulfuric acid, sewer gas, and stink damp.
Complete answer:
We need to know that copper sulphate, or copper(II) sulphate, is an inorganic substance having the chemical formula $CuS{O_4}$. The most common salt is the pentahydrate form ($CuS{O_4}$.\[5{H_2}O\]), which is brilliant blue. When hydrogen sulphide is passed through a blue copper sulphate solution, a black copper sulphide precipitate is formed, as well as sulphuric acid, which remains in the solution.
When hydrogen sulphide is passed through a blue copper sulphate solution, a black copper sulphide precipitate is produced, as well as sulphuric acid, which stays in the solution. The equation is as follows:
\[{H_2}S\left( g \right) + CuS{O_4}\;\left( {aq} \right) \to CuS\left( s \right) + {H_2}S{O_4}\left( {aq} \right)\]
Additional information:
A black colour precipitate forms in the test tube when hydrogen sulphide gas is fed through copper sulphate solution. It's a confirmatory test for sulphate ion presence in salt. Because hydrogen sulphide is somewhat denser than air, a combination of the two can be explosive. Sulfur dioxide and water are formed when hydrogen sulphide burns in oxygen with a blue flame. In general, hydrogen sulphide works as a reducing agent, particularly when a base is present.
Note:
We must note that the reaction of ${H_2}S$ gas when passed through copper sulphate solution is an example of double displacement reaction. Double-displacement reactions: This is also a displacement type of reaction which is represented by \[AB + CD \to AD + CB\]. The cation from one reactant displaces or replaces the cation from another reactant to form two new compounds.
Complete answer:
We need to know that copper sulphate, or copper(II) sulphate, is an inorganic substance having the chemical formula $CuS{O_4}$. The most common salt is the pentahydrate form ($CuS{O_4}$.\[5{H_2}O\]), which is brilliant blue. When hydrogen sulphide is passed through a blue copper sulphate solution, a black copper sulphide precipitate is formed, as well as sulphuric acid, which remains in the solution.
When hydrogen sulphide is passed through a blue copper sulphate solution, a black copper sulphide precipitate is produced, as well as sulphuric acid, which stays in the solution. The equation is as follows:
\[{H_2}S\left( g \right) + CuS{O_4}\;\left( {aq} \right) \to CuS\left( s \right) + {H_2}S{O_4}\left( {aq} \right)\]
Additional information:
A black colour precipitate forms in the test tube when hydrogen sulphide gas is fed through copper sulphate solution. It's a confirmatory test for sulphate ion presence in salt. Because hydrogen sulphide is somewhat denser than air, a combination of the two can be explosive. Sulfur dioxide and water are formed when hydrogen sulphide burns in oxygen with a blue flame. In general, hydrogen sulphide works as a reducing agent, particularly when a base is present.
Note:
We must note that the reaction of ${H_2}S$ gas when passed through copper sulphate solution is an example of double displacement reaction. Double-displacement reactions: This is also a displacement type of reaction which is represented by \[AB + CD \to AD + CB\]. The cation from one reactant displaces or replaces the cation from another reactant to form two new compounds.
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