Question

The reaction of slaked lime with $C{l_2}$ gas gives:
A. Only $CaOC{l_2}$
B. Only $CaC{l_2}$
C. A mixture of \[Ca{\left( {OCl} \right)_2},{\text{ }}Ca{\left( {OH} \right)_2},{\text{ }}CaC{l_2}{\text{ }}and{\text{ }}{H_2}O\]
D. quick lime

Answer 1

Hint: Slaked lime is referred to as \[Ca{\left( {OH} \right)_2}\]. Calcium hydroxide which is an inorganic compound and has a colorless crystal or is a white powder and is produced when quicklime chemically known as calcium oxide is mixed with water. It is known to have many common names slaked lime, builder’s lime, etc.

Complete answer:
When Slaked lime \[Ca{\left( {OH} \right)_2}\] is reacted​ with chlorine gives bleaching powder $Ca{(OCl)_2}$ is formed as a product, calcium chloride and water are produced as a by-product. The reaction for bleaching powder formation is as follows:
\[2Ca{\left( {OH} \right)_2} + 2C{l_2} \to Ca{\left( {OCl} \right)_2} + CaC{l_2} + 2{H_2}O.\]
So, the correct answer is Option C.

Additional Information: Bleaching powder also known as Calcium hypochlorite is an inorganic compound which has a chemical formula of $Ca{(OCl)_2}$ . And is one of the main Ingredients when it comes to commercial products such as bleaching powder or chlorine powder and is used for water treatment and as a bleaching agent even.
Calcium hypochlorite is also used as an oxidizing agent and thus finds some use in Organic chemistry, for example, it is used in cleaving glycols, α-hydroxycarboxylic acids, and keto acids which gives aldehydes or carboxylic acids as products. Calcium hypochlorite can also be used in the manufacturing of chloroform or also in the haloform reaction.

Note:
Calcium hypochlorite exists in two forms one is the dry calcium hypochlorite and the other is anhydrous hypochlorite, and instead of the above-mentioned products, the bleaching powder instead is a mixture of calcium hypochlorite $Ca{(OCl)_2}$, dibasic calcium hypochlorite with chemical formula \[C{a_3}{(OCl)_2}{(OH)_4}\] and dibasic calcium chloride with chemical formula \[C{a_3}C{l_2}{(OH)_4}\].