Question

(i)What is observed when a solution of potassium iodide is added to a solution of lead nitrate taken

in a tube?

(ii) What type of reaction is this?

(iii) Write a balanced chemical equation to represent the above reaction.​

Answer 1

$\huge\star\underbrace{\mathtt\red{⫷❥ᴀ᭄} \mathtt\green{n~ }\mathtt\blue{ §} \mathtt\purple{₩}\mathtt\orange{Σ} \mathtt\pink{R⫸}}\star\:$

(i) A yellow precipitate of lead iodide (PbI

2

) Is formed.

(ii) Precipitation reactior/Double displacement reaction.

(iii)

Leadnitrate

Pb(NO

3

)

2

(aq)

+

Potassiumiodide

2KI(aq)

→

Leadiodide(YellowPPt.)

PbI

2

(s)

+

Potassiumnitrate

2KNO

3

(aq)

Explanation:

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