Question

when potassium iodine solution is added to a solution of lead (ii) nitrate in a test tube , a precipitate is formed . 1 = WHAT IS THE NAME OF THE PRECIPITATE ! 2 = NAME THE COMPOUND PRECIPITATED ! 3 = WRITE A BALANCED EQUATION OF THIS REACTION ! 4 = WHAT TYPE OF REACTION IS THIS!

Answer 1

Given,

when potassium iodide solution is added to a solution of lead (II) nitrate in a test tube, a precipitate is formed.

To find,

1. the name of the precipitate.
2. write balanced chemical equation of reaction
3. type of reaction is this.

potassium iodide = KI

lead (II) nitrate = Pb(NO₃)₂

Pb(NO₃)₂ + 2KI ⇒PbI₂ ↓ + 2KNO₃

here it is clear that when one mole of lead(II) nitrate reacts with 2 moles of potassium iodide, one mole of lead iodide (precipitate) and 2 moles of potassium nitrate are formed.

now you can answer easily,

1. lead iodide (PbI₂)

2. Pb(NO₃)₂ + 2KI ⇒PbI₂ ↓ + 2KNO₃

3. This is double displacement reaction as you can see.