The reaction of water with slaked lime is ______reaction. Here calcium oxide and water react to form______?

Hint: Calcium hydroxide is also known as slaked lime and is white powder. It forms suspension when it is added into water.

Complete Step by Step solution:
As we are already familiar with term ”QuickLime” which is used for calcium oxide and the chemical formula for calcium oxide is $CaO$ .The chemical formula for slaked lime is $Ca\mathop {\left( {OH} \right)}\nolimits_2 $.
Quick lime i.e. Calcium oxide when exposed in atmosphere forms calcium carbonate on reaction with carbon dioxide and calcium hydroxide which is formed on absorbing moisture. The reactions are given below:
$
  CaO + \mathop H\nolimits_2 O \to Ca\mathop {\left( {OH} \right)}\nolimits_2 \\
  CaO + C\mathop O\nolimits_2 \to CaC\mathop O\nolimits_3 \\
$
When water is added to calcium oxide, it forms calcium hydroxide. During this process, hissing sound is observed and also a large amount of heat is evolved which converts water into steam. This process is termed as slaking of lime and fine powder obtained is called ‘slaked lime’. The reaction is:
$CaO + \mathop H\nolimits_2 O \to Ca\mathop {\left( {OH} \right)}\nolimits_2 $.
When slaked lime is added in water it turns into a suspension which is called “milk of lime”. After some time the solution becomes clear and is known as “lime water”.

So, on above arguments we can say that reaction of water with slaked lime is an ‘exothermic’ reaction. Also, from above it is cleared that calcium oxide and water react to form ‘calcium hydroxide or slaked lime’.

Note: It should be remembered that when carbon dioxide is passed through lime water, it turns milky due to formation of calcium carbonate. This reaction is used as a test to identify carbon dioxide gas. The reaction is:
$Ca\mathop {\left( {OH} \right)}\nolimits_2 + C\mathop O\nolimits_2 \to \mathop {CaC\mathop O\nolimits_3 \downarrow }\limits_{\left( {Milkiness} \right)} + \mathop H\nolimits_2 O$
When excess of $CO_2$ is passed in above reaction, the milkiness disappears due to the formation of soluble calcium bicarbonate. The reaction is:
$CaC\mathop O\nolimits_3 + C\mathop O\nolimits_2 + \mathop H\nolimits_2 O \to \mathop {Ca\mathop {\left( {HC\mathop O\nolimits_3 } \right)}\nolimits_2 }\limits_{\left( {So\operatorname{lub} le} \right)} $.