Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How is Calcium Sulphate produced from Calcium Oxide?
(A) $\text{ CaO + }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$
(B) $\text{ CaO + S}{{\text{O}}_{\text{2}}}\text{ }$
(C) $\text{ CaO + S}{{\text{O}}_{\text{3}}}\text{ }$
(D) $\text{ CaO + S }$

seo-qna
SearchIcon
Answer
VerifiedVerified
459.3k+ views
The neutralization reaction is a reaction between the acid and base to form salt and base. The metal oxides like $\text{ MO }$are basic. The metal oxides when reacting with the dilute acid like $\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$ the acid and base react to form a salt and water.

Complete step by step solution:
All the metals of the alkaline earth metal burn-in presence of the $\text{ }{{\text{O}}_{\text{2}}}\text{ }$ and form a metal oxide having the general formula$\text{ MO }$. The calcium is an alkaline earth metal that reacts with the oxygen and it is converted into the calcium oxide$\text{ CaO }$.
The neutralization reaction is as depicted below,
$\text{ Acid + Base }\to \text{ Salt + water }$
The acid is generally the specie that donates the proton and the base is the substance that donates the hydroxide however, the bases can be those which react with acid and form salt and water.
$\text{ MO + 2}{{\text{H}}^{\text{+}}}\text{ }\to \text{ }{{\text{M}}^{\text{+}}}\text{ + }{{\text{H}}_{\text{2}}}\text{O }$
The metal oxides are basic. The metal oxides react with the acid. In reaction with the acid, the metal oxides form salt and the water. The reaction between the acid and base to form a salt and water is called the neutralization reaction.
The metal hydroxides are also alkaline and these release the hydroxide ions in the solution. When the metal oxides are reacting with the water they form the strong bases. The calcium oxides when reacting with the water it forms the strong base
$\text{ CaO + }{{\text{H}}_{\text{2}}}\text{O }\to \text{ Ca(OH}{{\text{)}}_{\text{2}}}\text{ }$
This can be proven by the litmus paper test. The strong base like calcium hydroxides $\text{Ca(OH}{{\text{)}}_{\text{2}}}$ changes the red litmus paper to blue. This confirms that the $\text{Ca(OH}{{\text{)}}_{\text{2}}}$ is a base.
When the calcium oxide is in the water it is converted into a strong base$\text{Ca(OH}{{\text{)}}_{\text{2}}}$.
The metal oxide like $\text{ CaO }$reacts with the acid such as $\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$ to form the salt of calcium sulphate and the water. The reaction between the calcium oxide and the sulphuric acid is shown below:
$\text{ CaO + }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }\to \text{ CaS}{{\text{O}}_{\text{4}}}\text{ + }{{\text{H}}_{\text{2}}}\text{O }$
The calcium sulphate $\text{CaS}{{\text{O}}_{\text{4}}}$ is salt of calcium oxide (base) and the sulphuric acid$\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$.

So, the answe is option (B).

Additional information:
The non-metallic oxides like carbon dioxide $\text{ C}{{\text{O}}_{\text{2 }}}$ are acidic. The carbon dioxide when reacted with the base. When carbon dioxide is dissolved in the water, it forms carbonic acid.
$\text{ C}{{\text{O}}_{\text{2}}}\text{ + }{{\text{H}}_{\text{2}}}\text{O }\to \text{ }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$
The carbonic acid has a $\text{ pH }$ value equal to$\text{ 4}\text{.68 }$. The carbonic acid is acidic. Thus we know here that the non-metallic oxides are acidic.
The acid or basic nature can be proven by the litmus paper test.

Note: The calcium sulphate $\text{ CaS}{{\text{O}}_{\text{4}}}\text{ }$is a calcium hydrate. It is a white solid and barely soluble in water.it is well known as the plaster of Paris or gypsum. The plaster of Paris is heated to $\text{ 393 K }$ obtain the gypsum. The reaction is as follows,
$\text{ CaS}{{\text{O}}_{\text{4}}}\text{.2}{{\text{H}}_{\text{2}}}\text{O }\to \text{ CaS}{{\text{O}}_{\text{4}}}\text{ }\text{.}\dfrac{\text{1}}{\text{2}}{{\text{H}}_{\text{2}}}\text{O + 1}\dfrac{\text{1}}{\text{2}}{{\text{H}}_{\text{2}}}\text{O}\uparrow $