Question

Write the method of preparation for the following compounds.
i.$MgC{{l}_{2}}.6{{H}_{2}}O\text{ (from MgO)}$
ii.$CaO\text{ (from limestone }\left( CaC{{O}_{3}} \right)$
iii.$CaC{{O}_{3}}\text{ }\left( from\text{ CaO} \right)$
iv.$CuC{{O}_{3}}\text{ }\left( from\text{ Cu}{{\left( OH \right)}_{2}} \right)$
v.$Plaster\text{ of Paris }\left( CaS{{O}_{4}}.\tfrac{1}{2}{{H}_{2}}O \right)\text{ }\left[ from\text{ gypsum} \right]$

Answer 1

Hint: A compound is prepared when certain reactants combine in order to form a product. Preparation reaction is the characteristic feature of a compound and particular conditions such as temperature, heat, catalyst are necessary for their formation.

Complete answer:
a) Formation of $MgC{{l}_{2}}.6{{H}_{2}}O\text{ from MgO}$
Magnesium Oxide reacts with hydrochloric acid and water to form Magnesium chloride hexahydrate.
\[MgO+2HCl+5{{H}_{2}}O\to MgC{{l}_{2}}.6{{H}_{2}}O\]
The $6{{H}_{2}}O$is the water of crystallisation present in the compound.

b) Formation of $CaO\text{ (from limestone }\left( CaC{{O}_{3}} \right)$
Calcium oxide is prepared by the thermal decomposition of Calcium carbonate. Calcium carbonate is heated at ${{825}^{o}}C$, to get Calcium Oxide and Carbon-di-oxide.
\[CaC{{O}_{3}}\left( s \right)\text{ }\to \text{ }CaO\left( s \right)\text{ }+\text{ }C{{O}_{2}}\left( g \right)\]

c) Formation of $CaC{{O}_{3}}\text{ }\left( from\text{ CaO} \right)$
Preparing$CaC{{O}_{3}}$from CaO is a two-step process.
Firstly, water is added to calcium oxide to give calcium hydroxide.
$CaO\text{ + }{{\text{H}}_{2}}O\text{ }\to \text{ }$ \[CaO\left( s \right)\text{ }+\text{ }{{H}_{2}}O\text{ }\left( aq \right)\to Ca{{\left( OH \right)}_{2}}\left( aq \right)\]
Secondly, carbon-di-oxide is passed over the obtained calcium hydroxide to get calcium carbonate and water is released.
\[Ca{{\left( OH \right)}_{2}}~\left( aq \right)+\text{ }C{{O}_{2}}~\left( g \right)\to \text{ }CaC{{O}_{3}}\downarrow \left( aq \right)\text{ }+\text{ }{{H}_{2}}O\left( aq \right)\]
The calcium carbonate made is in the precipitate form.

d) Formation of $CuC{{O}_{3}}\text{ }\left( from\text{ Cu}{{\left( OH \right)}_{2}} \right)$
Copper (II) carbonate $(CuC{{O}_{3}}\text{) }$ is also known as cupric carbonate or copper carbonate. It is present in the basic form as Copper (II) carbonate-dihydroxide.
Copper hydroxide reacts with carbon dioxide to form copper carbonate –hydroxide and water.
\[\]\[Cu{{\left( OH \right)}_{2}}\left( aq \right)~+\text{ }C{{O}_{2}}\left( g \right)~\to \text{ }C{{u}_{2}}{{(OH)}_{2}}C{{O}_{3}}~\left( aq \right)+\text{ }{{H}_{2}}O\left( aq \right)\]
This copper carbonate –hydroxide is dehydrated to form copper carbonate.
\[C{{u}_{2}}{{(OH)}_{2}}C{{O}_{3}}~\left( aq \right)\to \text{ CuC}{{\text{O}}_{3}}+\text{ }{{\text{H}}_{2}}O\]
e) Formation of $Plaster\text{ of Paris }\left( CaS{{O}_{4}}.\tfrac{1}{2}{{H}_{2}}O \right)\text{ }\left[ from\text{ gypsum} \right]$
Gypsum is heated at ${{120}^{o}}C$ to form plaster of Paris.
\[\text{CaS}{{\text{O}}_{4}}.2{{H}_{2}}O\text{ }\to \text{ CaS}{{\text{O}}_{4}}.\tfrac{1}{2}{{H}_{2}}O\text{ }+\text{ }\tfrac{3}{2}{{H}_{2}}O\]
The chemical name for Plaster of Paris is calcium sulphate hemihydrate.

Note:
While writing the chemical equation, make sure to correctly balance the equation. Also, mention the state symbols of every compound on the reactant as well as product side.
Water of crystallisation- It is the number of molecules of water that are bonded chemically into the crystal structure.